\displaystyle \textbf{Question 1. }\text{The distance of a point }A\text{ from }x\text{-axis is }3\text{ units. Which of the following}
\displaystyle \text{cannot be coordinates of the point }A? \hspace{0.2cm}\text{[CBSE 2025]}
\displaystyle \text{(a) }(1,3) \qquad \text{(b) }(-3,-3)
\displaystyle \text{(c) }(-3,3) \qquad \text{(d) }(3,1)
\displaystyle \text{Answer:}
\displaystyle \textbf{(d)}
\displaystyle \text{Distance of a point }A\text{ from x-axis is }3\text{ units i.e. y-coordinate is }3\text{ or }-3.
\displaystyle \text{So, }(3,1)\text{ cannot be possible.}
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\displaystyle \textbf{Question 2. }P(x,y),Q(-2,-3)\text{ and }R(2,3)\text{ are the vertices of a right triangle}
\displaystyle PQR\text{ right angled at }P. \text{Find the relationship between }x\text{ and }y. \text{Hence, find all}
\displaystyle \text{possible values of }x\text{ for which }y=2. \hspace{0.2cm}\text{[CBSE 2025]}
\displaystyle \text{Answer:}
\displaystyle \text{As, }\triangle PQR\text{ is a right angled triangle, so by Pythagoras theorem,}\displaystyle RP^2+PQ^2=RQ^2
\displaystyle (x-2)^2+(y-3)^2+(x+2)^2+(y+3)^2=16+36
\displaystyle \Rightarrow x^2+4-4x+y^2+9-6y+x^2+4+4x+y^2+9+6y=52
\displaystyle \Rightarrow 2x^2+2y^2+26=52
\displaystyle \Rightarrow 2x^2+2y^2=26
\displaystyle \Rightarrow x^2+y^2=13
\displaystyle \Rightarrow x^2=13-y^2
\displaystyle \text{This is the relation between }x\text{ and }y.
\displaystyle \text{Now, we have to find all values of }x\text{ for which }y=2
\displaystyle x^2=13-4
\displaystyle x^2=9
\displaystyle x=\pm 3
\displaystyle \text{i.e. }x=+3,-3
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\displaystyle \textbf{Question 3. }\text{Find the coordinates of the point }C\text{ which lies on the line }AB
\displaystyle \text{produced such that }AC=2BC,\text{ where coordinates of points }A\text{ and }B\text{ are}
\displaystyle (-1,7)\text{ and }(4,-3)\text{ respectively.} \hspace{0.2cm}\text{[CBSE 2025]}
\displaystyle \text{Answer:}
\displaystyle \text{Given coordinates of }A\text{ and }B\text{ are }(-1,7)\text{ and }(4,-3)\text{ respectively. ATQ}
\displaystyle AC=2BC
\displaystyle \Rightarrow AB+BC=2BC
\displaystyle \Rightarrow AB:BC=1:1
\displaystyle \text{We have to find out coordinate of }C.\text{ By section formula,}
\displaystyle P(x,y)=\left(\frac{m_1x_2+m_2x_1}{m_1+m_2},\frac{m_1y_2+m_2y_1}{m_1+m_2}\right)
\displaystyle (4,-3)=\left(\frac{x-1}{2},\frac{y+7}{2}\right)
\displaystyle \text{On equating, we get}
\displaystyle 4=\frac{x-1}{2},\qquad -3=\frac{y+7}{2}
\displaystyle \Rightarrow x=9,\qquad y=-13
\displaystyle \text{So, coordinates of }C\text{ are }(9,-13).
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\displaystyle \textbf{Question 4. }P(x,y),Q(-2,-3)\text{ and }R(2,3)\text{ are the vertices of a right triangle}
\displaystyle PQR\text{ right angled at }P. \text{Find the relationship between }x\text{ and }y. \text{Hence, find all possible}
\displaystyle \text{values of }x\text{ for which }y=2. \hspace{0.2cm}\text{[CBSE 2025]}
\displaystyle \text{Answer:}
\displaystyle \text{As, }\triangle PQR\text{ is a right angled triangle, so by Pythagoras theorem,}\displaystyle RP^2+PQ^2=RQ^2
\displaystyle (x-2)^2+(y-3)^2+(x+2)^2+(y+3)^2=16+36
\displaystyle \Rightarrow x^2+4-4x+y^2+9-6y+x^2+4+4x+y^2+9+6y=52
\displaystyle \Rightarrow 2x^2+2y^2+26=52
\displaystyle \Rightarrow 2x^2+2y^2=26
\displaystyle \Rightarrow x^2+y^2=13
\displaystyle \Rightarrow x^2=13-y^2
\displaystyle \text{This is the relation between }x\text{ and }y.
\displaystyle \text{Now, we have to find all values of }x\text{ for which }y=2
\displaystyle x^2=13-4
\displaystyle x^2=9
\displaystyle x=\pm 3
\displaystyle \text{i.e. }x=+3,-3
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\displaystyle \textbf{Question 5. }\text{If the distance between the points }(3,-5)\text{ and }(x,-5)\text{ is }15\text{ units,}
\displaystyle \text{then the values of }x\text{ are:} \hspace{0.2cm}\text{[CBSE 2024]}
\displaystyle \text{(a) }12,-18 \qquad \text{(b) }-12,18
\displaystyle \text{(c) }18,5 \qquad \text{(d) }-9,-12
\displaystyle \text{Answer:}
\displaystyle \textbf{(b)}
\displaystyle \text{Distance between the points }(x_1,y_1)\text{ and }(x_2,y_2)\text{ is given by,}
\displaystyle \text{Distance}=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}
\displaystyle \text{Let }x_1=3,\ y_1=-5,\ x_2=x,\ y_2=-5
\displaystyle \therefore 15=\sqrt{(x-3)^2+(-5+5)^2}
\displaystyle \Rightarrow (x-3)^2=225\Rightarrow x-3=\pm 15
\displaystyle \Rightarrow x=18\ \text{or}\ x=-12
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\displaystyle \textbf{Question 6. }AD\text{ is a median of }\triangle ABC\text{ with vertices }A(5,-6),B(6,4)\text{ and}
\displaystyle C(0,0). \text{Length }AD\text{ is equal to:} \hspace{0.2cm}\text{[CBSE 2024]}
\displaystyle \text{(a) }\sqrt{68}\text{ units} \qquad \text{(b) }2\sqrt{15}\text{ units}
\displaystyle \text{(c) }\sqrt{101}\text{ units} \qquad \text{(d) }10\text{ units}
\displaystyle \text{Answer:}
\displaystyle \textbf{(a)}\displaystyle \text{Since, }AD\text{ is median of }\triangle ABC,\text{ then }D\text{ is mid-point of }BC.
\displaystyle \text{Coordinates of }D=\left(\frac{6+0}{2},\frac{4+0}{2}\right)\text{ i.e. }(3,2)
\displaystyle \text{Now, }AD=\sqrt{(3-5)^{2}+(2+6)^{2}}
\displaystyle =\sqrt{4+64}=\sqrt{68}\text{ units}
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\displaystyle \textbf{Question 7. }\text{The centre of a circle is at }(2,0). \text{If one end of a diameter is at }(6,0),
\displaystyle \text{then the other end is at:} \hspace{0.2cm}\text{[CBSE 2024]}
\displaystyle \text{(a) }(0,0) \qquad \text{(b) }(4,0) \qquad \text{(c) }(-2,0) \qquad \text{(d) }(-6,0)
\displaystyle \text{Answer:}
\displaystyle \textbf{(c)}\displaystyle \text{Let }(x,y)\text{ be the other end of the diameter.}
\displaystyle \text{Now, coordinates of centre are }(2,0)\text{ and one end of diameter is }(6,0).
\displaystyle \text{By using mid-point formula:}
\displaystyle 2=\frac{x+6}{2}\text{ and }0=\frac{y+0}{2}
\displaystyle \Rightarrow x=-2\text{ and }y=0
\displaystyle \text{So, coordinates of other end are }(-2,0).
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\displaystyle \textbf{Question 8. }\text{Find the ratio in which the point }\left(\frac{8}{5},y\right)\text{ divides the line segment}
\displaystyle \text{joining the points }(1,2)\text{ and }(2,3). \text{Also, find the value of }y. \hspace{0.2cm}\text{[CBSE 2024]}
\displaystyle \text{Answer:}
\displaystyle \text{Suppose the point }P\left(\frac{8}{5},y\right)\text{ divides the line segment joining the points }A(1,2)\text{ and }B(2,3)\text{ in the ratio of }k:1.
\displaystyle \text{We know that coordinates of a point }(x,y)\text{ that divides the line segment joining the points }(x_1,y_1)\text{ and }(x_2,y_2)\text{ in the ratio }m_1:m_2
\displaystyle \text{are given by}
\displaystyle x=\frac{m_1x_2+m_2x_1}{m_1+m_2},\qquad y=\frac{m_1y_2+m_2y_1}{m_1+m_2}
\displaystyle \text{So,}
\displaystyle \frac{8}{5}=\frac{k\times 2+1\times 1}{k+1}
\displaystyle \Rightarrow 10k+5=8k+8
\displaystyle \Rightarrow 2k=3\Rightarrow k=\frac{3}{2}
\displaystyle \therefore \text{Required ratio is }3:2.
\displaystyle \text{Also,}
\displaystyle y=\frac{k\times 3+1\times 2}{k+1}
\displaystyle =\frac{\frac{3}{2}\times 3+2}{\frac{3}{2}+1}
\displaystyle =\frac{13}{5}
\displaystyle \therefore \text{Value of }y\text{ is }\frac{13}{5}.
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\displaystyle \textbf{Question 9. }ABCD\text{ is a rectangle formed by the points }A(-1,-1),B(-1,6),C(3,6)
\displaystyle \text{and }D(3,-1). P,Q,R\text{ and }S\text{ are mid-points of sides }AB,BC,CD\text{ and }DA
\displaystyle \text{respectively. Show that diagonals of the quadrilateral }PQRS\text{ bisect each other.}
\displaystyle \hspace{0.2cm}\text{[CBSE 2024]}
\displaystyle \text{Answer:}
\displaystyle \text{We know that, coordinates of the mid-point of the line segment joining the points }(x_1,y_1)\text{ and }(x_2,y_2)\displaystyle \text{are given by } \left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right).
\displaystyle \text{Now, }P,Q,R\text{ and }S\text{ are the mid-points of }AB,BC,CD\text{ and }DA.
\displaystyle \text{Coordinates of }P=\left(\frac{-1-1}{2},\frac{6-1}{2}\right)=\left(-1,\frac{5}{2}\right)
\displaystyle \text{Coordinates of }Q=\left(\frac{-1+3}{2},\frac{6+6}{2}\right)=(1,6)
\displaystyle \text{Coordinates of }R=\left(\frac{3+3}{2},\frac{6-1}{2}\right)=\left(3,\frac{5}{2}\right)
\displaystyle \text{Coordinates of }S=\left(\frac{-1+3}{2},\frac{-1-1}{2}\right)=(1,-1)
\displaystyle \text{Coordinates of mid-point of diagonal }PR
\displaystyle =\left(\frac{3-1}{2},\frac{\frac{5}{2}+\frac{5}{2}}{2}\right)=\left(1,\frac{5}{2}\right)
\displaystyle \text{Coordinates of mid-point of diagonal }QS
\displaystyle =\left(\frac{1+1}{2},\frac{6-1}{2}\right)=\left(1,\frac{5}{2}\right)
\displaystyle \text{Since coordinates of mid-points of both diagonals coincide,}
\displaystyle \therefore \text{Diagonals of the quadrilateral }PQRS\text{ bisect each other.}
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\displaystyle \textbf{Question 10. }\text{Partha, a software engineer, lives in Jerusalem for his work. He lives}
\displaystyle \text{in the most convenient area of the city from where bank, hospital, post office and}
\displaystyle \text{supermarket can be easily accessed. In the graph, the bank is plotted as }A(9,5),
\displaystyle \text{hospital as }B(-3,-1)\text{ and supermarket as }C(5,-5)\text{ such that }A,B,C\text{ form a triangle.}
\displaystyle \text{Based on the above given information, answer the following questions:} \hspace{0.2cm}\text{[CBSE 2024(C)]}
\displaystyle \text{(i) Find the distance between the bank and the hospital.}
\displaystyle \text{(ii) In between the bank and the supermarket, there is a post office plotted at }E
\displaystyle \text{which is their mid-point. Find the coordinates of }E.
\displaystyle \text{(iii) (a) In between the hospital and the supermarket, there is a bus stand plotted}
\displaystyle \text{as }D,\text{ which is their mid-point. If Partha wants to reach the bus stand from the}
\displaystyle \text{bank, then how much distance does he need to cover?}
\displaystyle \text{OR}
\displaystyle \text{(iii) (b) }P\text{ and }Q\text{ are two different garment shops lying between the bank and the}
\displaystyle \text{hospital, such that }BP=PQ=QA. \text{If the coordinates of }P\text{ and }Q\text{ are }(1,a)
\displaystyle \text{and }(b,3)\text{ respectively, then find the values of }a\text{ and }b.
\displaystyle \text{Answer:}
\displaystyle \text{(i) Distance between bank and hospital}
\displaystyle =\sqrt{(-3-9)^2+(-1-5)^2}
\displaystyle =\sqrt{180}
\displaystyle =6\sqrt{5}\text{ units}
\displaystyle \text{(ii) Coordinates of }E\text{ are}
\displaystyle \left(\frac{9+5}{2},\frac{5+(-5)}{2}\right)=(7,0)
\displaystyle \text{(iii) (a) Coordinates of }D\text{ are}
\displaystyle \left(\frac{-3+5}{2},\frac{-1+(-5)}{2}\right)=(1,-3)
\displaystyle \text{Distance Partha need to cover}
\displaystyle =\sqrt{(9-1)^2+(5-(-3))^2}
\displaystyle =\sqrt{128}
\displaystyle =8\sqrt{2}\text{ units}
\displaystyle \text{OR}
\displaystyle \text{(iii) (b) }P\text{ is midpoint of }BQ
\displaystyle a=\frac{-1+3}{2}=1
\displaystyle \text{and }Q\text{ is midpoint of }PA
\displaystyle b=\frac{1+9}{2}=5
\displaystyle \therefore (a,b)=(1,5).
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\displaystyle \textbf{Question 11. }\text{The distance of the point }(-1,7)\text{ from }x\text{-axis is:} \hspace{0.2cm}\text{[CBSE 2023]}
\displaystyle \text{(a) }-1 \qquad \text{(b) }7 \qquad \text{(c) }6 \qquad \text{(d) }\sqrt{50}
\displaystyle \text{Answer:}
\displaystyle \textbf{(b)}
\displaystyle \text{Distance of the point }(-1,7)\text{ from x-axis}=7\text{ units.}
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\displaystyle \textbf{Question 12. }\text{The distance of the point }(-6,8)\text{ from origin is:} \hspace{0.2cm}\text{[CBSE 2023]}
\displaystyle \text{(a) }6 \qquad \text{(b) }-6 \qquad \text{(c) }8 \qquad \text{(d) }10
\displaystyle \text{Answer:}
\displaystyle \textbf{(d)}
\displaystyle \text{Distance of the point }(-6,8)\text{ from origin}
\displaystyle =\sqrt{(-6)^2+(8)^2}=10
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\displaystyle \textbf{Question 13. }\text{Show that the four points }A(0,-1),B(6,7),C(-2,3)\text{ and }D(8,3)
\displaystyle \text{are the vertices of a rectangle.} \hspace{0.2cm}\text{[CBSE 2023(C)]}
\displaystyle \text{Answer:}
\displaystyle \textbf{(d)}
\displaystyle \text{As distance }d\text{ between points }(x_1,y_1)\text{ and }(x_2,y_2)\text{ is}
\displaystyle d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}
\displaystyle \text{So,}
\displaystyle AB=\sqrt{(6-0)^2+(7+1)^2}=10\text{ units}
\displaystyle BC=\sqrt{(-2-6)^2+(3-7)^2}=4\sqrt{5}\text{ units}
\displaystyle CD=\sqrt{(8+2)^2+(3-3)^2}=10\text{ units}
\displaystyle DA=\sqrt{(0-8)^2+(-1-3)^2}=4\sqrt{5}\text{ units}
\displaystyle AC=\sqrt{(-2-0)^2+(3+1)^2}=2\sqrt{5}\text{ units}
\displaystyle BD=\sqrt{(8-6)^2+(3-7)^2}=2\sqrt{5}\text{ units}
\displaystyle \text{In }\square ABCD,\text{ we have }AB=CD,\ BC=DA\text{ and }AC=BD.
\displaystyle \text{So, }ABCD\text{ is a rectangle.}
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\displaystyle \textbf{Question 14. }\text{If }P\text{ is the mid-point of the line segment forming the points}
\displaystyle A(-2,8)\text{ and }B(-6,-4),\text{ then the coordinates of }P\text{ are:} \hspace{0.2cm}\text{[CBSE 2023(C)]}
\displaystyle \text{(a) }(-4,2) \qquad \text{(b) }(2,-4)
\displaystyle \text{(c) }(6,8) \qquad \text{(d) }(-6,8)
\displaystyle \text{Answer:}
\displaystyle \textbf{(a)}
\displaystyle \text{We know that coordinates of mid-point of the line-segment joining the points }(x_{1},y_{1})\text{ and }(x_{2},y_{2})
\displaystyle \text{is }\left(\frac{x_{1}+x_{2}}{2},\frac{y_{1}+y_{2}}{2}\right)
\displaystyle \therefore \text{Coordinates of }P=\left(\frac{-2-6}{2},\frac{8-4}{2}\right)=(-4,2)
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\displaystyle \textbf{Question 15. }\text{Show that the points }A(6,4),B(5,-2)\text{ and }C(7,-2)\text{ are the vertices}
\displaystyle \text{of an isosceles triangle. Also, find the length of the median through point }A.
\displaystyle \hspace{0.2cm}\text{[CBSE 2023(C)]}
\displaystyle \text{Answer:}
\displaystyle \text{As, distance }(d)\text{ between points }(x_1,y_1)\text{ and }(x_2,y_2)\text{ is given by,}
\displaystyle d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\displaystyle AB=\sqrt{(5-6)^2+(-2-4)^2}=\sqrt{37}\text{ units}
\displaystyle BC=\sqrt{(7-5)^2+(-2+2)^2}=2\text{ units}
\displaystyle CA=\sqrt{(6-7)^2+(4+2)^2}=\sqrt{37}\text{ units}
\displaystyle \text{In }\triangle ABC,\text{ we have }AB=AC,\text{ then }\triangle ABC\text{ is an isosceles triangle.}
\displaystyle \text{Let }AD\text{ be median to side }BC.
\displaystyle \text{Now, }D\text{ is the mid-point of }BC.
\displaystyle \text{So, coordinates of point }D=\left(\frac{5+7}{2},\frac{-2-2}{2}\right)=(6,-2)
\displaystyle \text{Now, }AD=\sqrt{(6-6)^2+(-2-4)^2}=6\text{ units}.
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\displaystyle \textbf{Question 16. }\text{Find the ratio in which the point }(-1,k)\text{ divides the line segment}
\displaystyle \text{joining the points }(-3,10)\text{ and }(6,-8). \text{Hence, find the value of }k.
\displaystyle \hspace{0.2cm}\text{[CBSE 2023(C)]}
\displaystyle \text{Answer:}
\displaystyle \text{Suppose the point }(-1,k)\text{ divides the line segment joining the points }(-3,10)\text{ and }(6,-8)\text{ in the ratio }\lambda:1.
\displaystyle \text{By section formula,}
\displaystyle -1=\frac{\lambda\times 6+1\times(-3)}{\lambda+1}\quad \text{and}\quad k=\frac{\lambda\times(-8)+1\times 10}{\lambda+1}
\displaystyle \text{Now, consider the equation,}
\displaystyle -1=\frac{\lambda\times 6+1\times(-3)}{\lambda+1}
\displaystyle \Rightarrow 6\lambda-3=-\lambda-1
\displaystyle \Rightarrow \lambda=\frac{2}{7}
\displaystyle \text{So, required ratio is }2:7.
\displaystyle \text{Consider the equation,}
\displaystyle k=\frac{\lambda\times(-8)+10}{\lambda+1}
\displaystyle \Rightarrow k=\frac{\frac{2}{7}\times(-8)+10}{\frac{2}{7}+1}
\displaystyle \Rightarrow k=6
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\displaystyle \textbf{Question 17. }\text{Jagdish has a field which is in the shape of a right angled triangle }AQC.
\displaystyle \text{He wants to leave a space in the form of a square }PQRS\text{ inside the field for}
\displaystyle \text{growing wheat and the remaining for growing vegetables (as shown in the figure).}
\displaystyle \text{In the field, there is a pole marked as }O.
\displaystyle \text{Based on the above information, answer the following questions:} \hspace{0.2cm}\text{[CBSE 2023]}
\displaystyle \text{(i) Taking }O\text{ as origin, coordinates of }P\text{ are }(-200,0)\text{ and of }Q\text{ are }(200,0).
\displaystyle PQRS\text{ being a square, what are the coordinates of }R\text{ and }S?
\displaystyle \text{(ii) What is the area of square }PQRS?
\displaystyle \text{OR}
\displaystyle \text{What is the length of diagonal }PR\text{ in square }PQRS?
\displaystyle \text{(iii) If }S\text{ divides }CA\text{ in the ratio }k:1,\text{ what is the value of }k,\text{ where point }A
\displaystyle \text{is }(200,800)?
\displaystyle \text{Answer:}
\displaystyle \text{(i) Coordinates of }P(-200,0),\ Q(200,0),\ R(200,400)
\displaystyle S(-200,400)\qquad (\because PQRS\text{ is a square})
\displaystyle \text{(ii) Area of square }PQRS=\text{side}^2
\displaystyle PQ=OP+OQ=200+200=400
\displaystyle \therefore \text{Area of square }PQRS=400^2=160000\text{ sq units}
\displaystyle \text{OR}
\displaystyle \text{Length of diagonal }PR=\sqrt{2}\times\text{side}
\displaystyle =\sqrt{2}\times400=400\sqrt{2}\text{ units}
\displaystyle \text{(iii) Coordinates of }S=(-200,400).
\displaystyle \text{Let }S\text{ divide }AC\text{ in the ratio }k:1.
\displaystyle -200=\frac{k\times200+1\times(-600)}{k+1}
\displaystyle -200k-200=200k-600
\displaystyle 400=400k
\displaystyle k=1.
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\displaystyle \textbf{Question 18. }\text{If }A(3,\sqrt{3}),\ B(0,0)\text{ and }C(3,k)\text{ are the three vertices of an}
\displaystyle \text{equilateral triangle }ABC,\text{ then the value of }k\text{ is} \hspace{0.2cm}\text{[CBSE 2021]}
\displaystyle \text{(a) }2 \qquad \text{(b) }-3 \qquad \text{(c) }-\sqrt{3} \qquad \text{(d) }-\sqrt{2}
\displaystyle \text{Answer:}
\displaystyle \textbf{(c)}
\displaystyle \text{Since }ABC\text{ is an equilateral triangle, then}
\displaystyle AB=BC=CA
\displaystyle \Rightarrow AB=BC
\displaystyle \Rightarrow \sqrt{(0-3)^2+(0-\sqrt{3})^2}=\sqrt{(3-0)^2+(k-0)^2}
\displaystyle \Rightarrow 12=k^2+9\Rightarrow k=\pm\sqrt{3}
\displaystyle \text{So, }k=-\sqrt{3}\text{ or }\sqrt{3}
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\displaystyle \textbf{Question 19. }\text{If }A(4,-2),\ B(7,-2)\text{ and }C(7,9)\text{ are the vertices of a }\triangle ABC,
\displaystyle \text{then }\triangle ABC\text{ is} \hspace{0.2cm}\text{[CBSE 2021]}
\displaystyle \text{(a) equilateral triangle}
\displaystyle \text{(b) isosceles triangle}
\displaystyle \text{(c) right angled triangle}
\displaystyle \text{(d) isosceles right angled triangle}
\displaystyle \text{Answer:}
\displaystyle \textbf{(c)}
\displaystyle AB=\sqrt{(7-4)^2+(-2+2)^2}=\sqrt{9}=3\text{ units}
\displaystyle BC=\sqrt{(7-7)^2+(9+2)^2}=11\text{ units}
\displaystyle CA=\sqrt{(4-7)^2+(-2-9)^2}=\sqrt{130}\text{ units}
\displaystyle \text{Now, }CA^2=130
\displaystyle AB^2+BC^2=9+121=130
\displaystyle \text{Since }AC^2=AB^2+BC^2,\text{ by converse of Pythagoras theorem,}
\displaystyle \triangle ABC\text{ is right angle triangle with }\angle B=90^\circ.
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\displaystyle \textbf{Question 20. }\text{Three vertices of a parallelogram }ABCD\text{ are }A(1,4),B(-2,3)
\displaystyle \text{and }C(5,8). \text{The ordinate of the fourth vertex }D\text{ is} \hspace{0.2cm}\text{[CBSE 2021]}
\displaystyle \text{(a) }8 \qquad \text{(b) }9 \qquad \text{(c) }7 \qquad \text{(d) }6
\displaystyle \text{Answer:}
\displaystyle \textbf{(b)}
\displaystyle \text{Let the coordinates of }D\text{ be }(m,n).
\displaystyle \text{As, diagonals of parallelogram bisect each other, then}
\displaystyle \text{coordinates of mid-point of }BD=\text{coordinates of mid-point of }AC.
\displaystyle \Rightarrow \left(\frac{-2+m}{2},\frac{3+n}{2}\right)=\left(\frac{1+5}{2},\frac{4+8}{2}\right)
\displaystyle \Rightarrow -2+m=6\text{ and }3+n=12
\displaystyle \Rightarrow m=8\text{ and }n=9
\displaystyle \text{So, ordinate of }D\text{ is }9.
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\displaystyle \textbf{Question 21. }\text{The ratio in which the line }3x+y-9=0\text{ divides the line segment}
\displaystyle \text{joining the points }(1,3)\text{ and }(2,7)\text{ is} \hspace{0.2cm}\text{[CBSE 2021]}
\displaystyle \text{(a) }3:2 \qquad \text{(b) }2:3 \qquad \text{(c) }3:4 \qquad \text{(d) }4:3
\displaystyle \text{Answer:}
\displaystyle \textbf{(c)}
\displaystyle \text{Let the required ratio be }k:1.\text{ Let }P(\alpha,\beta)\text{ be the point of division.}
\displaystyle \text{Using section formula,}
\displaystyle \alpha=\frac{2k+1}{k+1};\ \beta=\frac{7k+3}{k+1}
\displaystyle \text{Now, }P\text{ lies on }3x+y-9=0,\text{ then it must satisfy it.}
\displaystyle 3\alpha+\beta=9
\displaystyle \Rightarrow 3\left(\frac{2k+1}{k+1}\right)+\left(\frac{7k+3}{k+1}\right)=9
\displaystyle \Rightarrow 13k+6=9k+9
\displaystyle \Rightarrow k=\frac{3}{4}
\displaystyle \text{So, ratio is }3:4.
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\displaystyle \textbf{Question 22. }\text{The line segment joining the points }P(-3,2)\text{ and }Q(5,7)\text{ is divided}
\displaystyle \text{by the }y\text{-axis in the ratio} \hspace{0.2cm}\text{[CBSE 2021]}
\displaystyle \text{(a) }3:1 \qquad \text{(b) }3:4
\displaystyle \text{(c) }3:2 \qquad \text{(d) }3:5
\displaystyle \text{Answer:}
\displaystyle \textbf{(d)}
\displaystyle \text{Let }R(0,a)\text{ be the point of division and let }k:1\text{ be the required ratio.}
\displaystyle \text{Using section formula,}
\displaystyle 0=\frac{k\times 5+1\times(-3)}{k+1}\Rightarrow k=\frac{3}{5}
\displaystyle \text{So, required ratio is }3:5.
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\displaystyle \textbf{Question 23. }\text{The base }BC\text{ of an equilateral }\triangle ABC\text{ lies on the }y\text{-axis.}
\displaystyle \text{The coordinates of }C\text{ are }(0,-3). \text{If the origin is the mid-point of the base }BC,
\displaystyle \text{what are the coordinates of }A\text{ and }B? \hspace{0.2cm}\text{[CBSE 2021]}
\displaystyle \text{(a) }A(\sqrt{3},0),B(0,3)
\displaystyle \text{(b) }A(\pm3\sqrt{3},0),B(3,0)
\displaystyle \text{(c) }A(\pm3\sqrt{3},0),B(0,3)
\displaystyle \text{(d) }A(-\sqrt{3},0),B(3,0)
\displaystyle \text{Answer:}
\displaystyle \textbf{(c)}
\displaystyle \text{Let the coordinates of }A\text{ and }B\text{ be }(x_1,y_1)\text{ and }(x_2,y_2).
\displaystyle \text{Now, }O(0,0)\text{ is mid-point of }BC.
\displaystyle \text{So, }(0,0)=\left(\frac{x_2+0}{2},\frac{y_2-3}{2}\right)
\displaystyle \Rightarrow (x_2,y_2)=(0,3)=\text{coordinates of }B.
\displaystyle \text{Now, }OA\text{ will be perpendicular bisector of }BC.
\displaystyle \text{So, }A\text{ will lie on x-axis. Hence, }(x_1,0)\text{ be the coordinates of }A.
\displaystyle \text{Since }ABC\text{ is an equilateral triangle, }AB=BC.
\displaystyle \Rightarrow \sqrt{(x_1-0)^2+(0-3)^2}=6
\displaystyle \Rightarrow x_1=\pm 3\sqrt{3}
\displaystyle \text{So, coordinates of }A=(\pm 3\sqrt{3},0).
\\

\displaystyle \textbf{Question 24. }\text{The distance between the points }(a\cos\theta+b\sin\theta,0)\text{ and}
\displaystyle (0,a\sin\theta-b\cos\theta)\text{ is} \hspace{0.2cm}\text{[CBSE 2020]}
\displaystyle \text{(a) }a^{2}+b^{2} \qquad \text{(b) }a^{2}-b^{2}
\displaystyle \text{(c) }\sqrt{a^{2}+b^{2}} \qquad \text{(d) }\sqrt{a^{2}-b^{2}}
\displaystyle \text{Answer:}
\displaystyle \textbf{(c)}
\displaystyle \text{As distance }
\displaystyle \sqrt{\left[(a\cos\theta+b\sin\theta-0)^2+(0-a\sin\theta+b\cos\theta)^2\right]}
\displaystyle =\sqrt{\left[a^2\cos^2\theta+b^2\sin^2\theta+2ab\sin\theta\cos\theta\right]}
\displaystyle \qquad\sqrt{\left[a^2\sin^2\theta+b^2\cos^2\theta-2ab\sin\theta\cos\theta\right]}
\displaystyle =\sqrt{a^2(\cos^2\theta+\sin^2\theta)+b^2(\sin^2\theta+\cos^2\theta)}
\displaystyle =\sqrt{a^2+b^2}
\\

\displaystyle \textbf{Question 25. }\text{If the point }P(k,0)\text{ divides the line segment joining the points}
\displaystyle A(2,-2)\text{ and }B(-7,4)\text{ in the ratio }1:2,\text{ then the value of }k\text{ is}
\displaystyle \hspace{0.2cm}\text{[CBSE 2020]}
\displaystyle \text{(a) }1 \qquad \text{(b) }2 \qquad \text{(c) }-2 \qquad \text{(d) }-1
\displaystyle \text{Answer:}
\displaystyle \textbf{(d)}
\displaystyle \text{For points }A(2,-2),\ P(k,0)\text{ and }B(-7,4),
\displaystyle \text{ratio }AP:PB=1:2
\displaystyle k=\frac{(-7)\times 1+2\times 2}{1+2}=\frac{-7+4}{3}=-1
\\

\displaystyle \textbf{Question 26. }\text{Write the coordinates of a point }P\text{ on }x\text{-axis which is equidistant from}
\displaystyle \text{the points }A(-2,0)\text{ and }B(6,0). \hspace{0.2cm}\text{[CBSE 2019]}
\displaystyle \text{Answer:}
\displaystyle \text{Let coordinates of }P\text{ be }(x,0).
\displaystyle \text{ATQ}\qquad AP=BP
\displaystyle \sqrt{(x+2)^2+(0-0)^2}=\sqrt{(x-6)^2+(0-0)^2}
\displaystyle \Rightarrow (x+2)^2=(x-6)^2
\displaystyle \Rightarrow x^2+4+4x=x^2+36-12x
\displaystyle \Rightarrow 16x=32\Rightarrow x=2
\displaystyle \therefore \text{Coordinates of point }P\text{ are }(2,0).
\\

\displaystyle \textbf{Question 27. }\text{Find the point on }y\text{-axis which is equidistant from the points }(5,-2)
\displaystyle \text{and }(-3,2). \hspace{0.2cm}\text{[CBSE 2019]}
\displaystyle \text{Answer:}
\displaystyle \text{Let point on y-axis be }(0,a).
\displaystyle \text{Now, distance of this point from }(5,-2)\text{ is equal to distance from }(-3,2),
\displaystyle \text{i.e., }\sqrt{5^2+(-2-a)^2}=\sqrt{(3)^2+(a-2)^2}
\displaystyle \text{Squaring and simplifying, we get}
\displaystyle 25+4+a^2+4a=9+a^2+4-4a
\displaystyle \Rightarrow 8a=-16
\displaystyle \Rightarrow a=-2
\\

\displaystyle \textbf{Question 28. }\text{Find the coordinates of a point }A,\text{ where }AB\text{ is a diameter of a circle}
\displaystyle \text{whose centre is }(2,-3)\text{ and }B\text{ is the point }(1,4). \hspace{0.2cm}\text{[CBSE 2019]}
\displaystyle \text{Answer:}
\displaystyle AB\text{ is diameter of the circle.}\displaystyle \text{Let }C\text{ be centre of circle, coordinates of }C\text{ are }(2,-3).
\displaystyle \text{Here, }C\text{ is mid-point of }AB\text{ (diameter).}
\displaystyle \text{Let coordinates of }A\text{ be }(x,y).
\displaystyle \therefore \frac{x+1}{2}=2\quad \text{and}\quad \frac{y+4}{2}=-3
\displaystyle \Rightarrow x+1=4\quad \text{and}\quad y+4=-6
\displaystyle \Rightarrow x=3\quad \text{and}\quad y=-10
\displaystyle \therefore \text{The coordinates of }A\text{ are }(3,-10).
\\

\displaystyle \textbf{Question 29. }\text{Find the ratio in which the segment joining the points }(1,-3)\text{ and }(4,5)
\displaystyle \text{is divided by }x\text{-axis? Also find the coordinates of this point on }x\text{-axis.} \hspace{0.2cm}\text{[CBSE 2019]}
\displaystyle \text{Answer:}
\displaystyle \text{Let x-axis divides }AB\text{ in the ratio }k:1\text{ at point }C(x,0)\text{ where coordinates of } \\ A\text{ and }B\text{ are }(1,-3)\text{ and }(4,5).
\displaystyle \text{Using section formula,}
\displaystyle x=\frac{4k+1}{k+1}
\displaystyle \text{and}
\displaystyle 0=\frac{5k-3}{k+1}
\displaystyle \Rightarrow 5k=3\Rightarrow k=\frac{3}{5}
\displaystyle \therefore C(x,0)\text{ divides }AB\text{ in the ratio }3:5.
\displaystyle \text{Also,}
\displaystyle x=\frac{4\left(\frac{3}{5}\right)+1}{\frac{3}{5}+1}=\frac{12+5}{3+5}=\frac{17}{8}
\displaystyle \therefore \text{Coordinates of }C\text{ are }\left(\frac{17}{8},0\right).
\\

\displaystyle \textbf{Question 30. }\text{The line segment joining the points }A(2,1)\text{ and }B(5,-8)\text{ is trisected}
\displaystyle \text{at the points }P\text{ and }Q\text{ such that }P\text{ is nearer to }A. \text{If }P\text{ also lies on the line}
\displaystyle \text{given by }2x-y+k=0,\text{ find the value of }k. \hspace{0.2cm}\text{[CBSE 2019]}
\displaystyle \text{Answer:}
\displaystyle \text{Since point }P\text{ and }Q\text{ trisect }AB,\text{ then }PA:PB=1:2.
\displaystyle \text{Coordinates of }P\text{ are}
\displaystyle x=\frac{5\times1+2\times2}{1+2}=\frac{5+4}{3}=3
\displaystyle \text{and}
\displaystyle y=\frac{-8\times1+1\times2}{1+2}=\frac{-8+2}{3}=-2
\displaystyle \text{Now, }P\text{ lies on }2x-y+k=0
\displaystyle \text{On putting values of }x\text{ and }y,\text{ we get}
\displaystyle 2\times3+2+k=0
\displaystyle \Rightarrow k=-8
\\

\displaystyle \textbf{Question 31. }\text{Find the linear relation between }x\text{ and }y\text{ such that }P(x,y)\text{ is}
\displaystyle \text{equidistant from the points }A(1,4)\text{ and }B(-1,2). \hspace{0.2cm}\text{[CBSE 2018(C)]}
\displaystyle \text{Answer:}
\displaystyle PA=PB
\displaystyle \sqrt{(x-1)^2+(y-4)^2}=\sqrt{(x+1)^2+(y-2)^2}
\displaystyle \text{Squaring both sides, we get}
\displaystyle (x-1)^2+(y-4)^2=(x+1)^2+(y-2)^2
\displaystyle \Rightarrow x^2-2x+1+y^2-8y+16=x^2+2x+1+y^2-4y+4
\displaystyle \Rightarrow -4x-4y=-12
\displaystyle \Rightarrow x+y=3
\\

\displaystyle \textbf{Question 32. }A(5,1),B(1,5)\text{ and }C(-3,-1)\text{ are the vertices of }\triangle ABC.
\displaystyle \text{Find the length of median }AD. \hspace{0.2cm}\text{[CBSE 2018(C)]}
\displaystyle \text{Answer:}
\displaystyle AD\text{ is median of }\triangle ABC,\text{ so }D\text{ is mid-point of }BC.\displaystyle \text{Coordinates of }D\text{ are:}
\displaystyle \left(\frac{1-3}{2},\frac{5-1}{2}\right)=\left(\frac{-2}{2},\frac{4}{2}\right)=(-1,2)
\displaystyle \text{Length of }AD=\sqrt{(5+1)^2+(1-2)^2}
\displaystyle =\sqrt{36+1}
\displaystyle =\sqrt{37}\text{ units.}
\\

\displaystyle \textbf{Question 33. }\text{If }A(-2,1),B(a,0),C(4,b)\text{ and }D(1,2)\text{ are the vertices}
\displaystyle \text{of a parallelogram }ABCD,\text{ find the values of }a\text{ and }b. \text{Hence find the lengths}
\displaystyle \text{of its sides.} \hspace{0.2cm}\text{[CBSE 2018]}
\displaystyle \text{Answer:}
\displaystyle \text{We know, the diagonals of a parallelogram bisect each other at a point. Let they bisect at }O.
\displaystyle \text{For diagonal }AC,\displaystyle \text{Coordinates of mid-point }O=\left(\frac{-2+4}{2},\frac{1+b}{2}\right)=\left(1,\frac{1+b}{2}\right)
\displaystyle \text{Also, for diagonal }BD,
\displaystyle \text{Coordinates of mid-point }O=\left(\frac{1+a}{2},\frac{2+0}{2}\right)=\left(\frac{1+a}{2},1\right)
\displaystyle \text{So,}
\displaystyle 1=\frac{1+a}{2}
\displaystyle \Rightarrow a=2-1=1
\displaystyle \text{Also,}
\displaystyle \frac{1+b}{2}=1
\displaystyle \Rightarrow b=1
\displaystyle \therefore a=1,\ b=1
\displaystyle \text{Length of side }AB=\sqrt{(a-(-2))^2+(0-1)^2}
\displaystyle =\sqrt{(1+2)^2+(-1)^2}
\displaystyle =\sqrt{10}\text{ units}
\displaystyle \text{Length of side }BC=\sqrt{(4-a)^2+(b-0)^2}
\displaystyle =\sqrt{(4-1)^2+(1-0)^2}
\displaystyle =\sqrt{10}\text{ units}
\displaystyle \text{Length of side }CD=\sqrt{(1-4)^2+(2-1)^2}
\displaystyle =\sqrt{9+1}=\sqrt{10}\text{ units}
\displaystyle \text{Length of side }AD=\sqrt{(1-(-2))^2+(2-1)^2}
\displaystyle =\sqrt{9+1}=\sqrt{10}\text{ units}
\displaystyle \therefore AB=BC=CD=DA=\sqrt{10}\text{ units.}
\\

\displaystyle \textbf{Question 34. }\text{If the distance between the points }(4,k)\text{ and }(1,0)\text{ is }5,\text{ then what can}
\displaystyle \text{be the possible values of }k? \hspace{0.2cm}\text{[CBSE 2017]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }A(4,k)\text{ and }B(1,0)\text{ be the given points.}
\displaystyle \text{Now,}\qquad AB=5
\displaystyle \Rightarrow \sqrt{(1-4)^2+(0-k)^2}=5\qquad [\text{Distance formula}]
\displaystyle \Rightarrow 9+k^2=25
\displaystyle \Rightarrow k^2=16
\displaystyle \Rightarrow k=\pm 4
\\

\displaystyle \textbf{Question 34. }\text{If the distances of }P(x,y)\text{ from }A(5,1)\text{ and }B(-1,5)\text{ are equal,}
\displaystyle \text{then prove that }3x=2y. \hspace{0.2cm}\text{[CBSE 2017]}
\displaystyle \text{Answer:}
\displaystyle PA=PB\qquad (\text{Given})
\displaystyle \Rightarrow \sqrt{(5-x)^2+(1-y)^2}=\sqrt{(-1-x)^2+(5-y)^2}\qquad [\text{Distance formula}]
\displaystyle \Rightarrow \sqrt{25+x^2-10x+1+y^2-2y}=\sqrt{1+x^2+2x+25+y^2-10y+26}
\displaystyle \Rightarrow x^2+y^2-10x-2y+26=x^2+y^2+2x-10y+26
\displaystyle \Rightarrow 12x=8y
\displaystyle \Rightarrow 3x=2y
\\

\displaystyle \textbf{Question 36. }\text{If the point }(x,y)\text{ is equidistant from the points }(a+b,b-a)\text{ and}
\displaystyle (a-b,a+b),\text{ prove that }bx=ay. \hspace{0.2cm}\text{[CBSE 2017(C)]}
\displaystyle \text{Answer:}
\displaystyle \text{Consider that point }P(x,y)\text{ is equidistant from }A(a+b,b-a)\text{ and }B(a-b,a+b)
\displaystyle PA=PB
\displaystyle \sqrt{[x-(a+b)]^2+[y-(b-a)]^2}=\sqrt{[x-(a-b)]^2+[y-(a+b)]^2}
\displaystyle \text{Squaring both sides,}
\displaystyle x^2+(a+b)^2-2(a+b)x+y^2+(b-a)^2-2(b-a)y
\displaystyle =x^2+(a-b)^2-2(a-b)x+y^2+(a+b)^2-2y(a+b)
\displaystyle \Rightarrow -2(a+b)x-2(b-a)y=-2(a-b)x-2(a+b)y
\displaystyle \Rightarrow (a+b)x+(b-a)y=(a-b)x+(a+b)y
\displaystyle \Rightarrow (a+b-a+b)x=(a+b-b+a)y
\displaystyle \Rightarrow 2bx=2ay
\displaystyle \Rightarrow bx=ay
\\

\displaystyle \textbf{Question 37. }\text{Show that }\triangle ABC,\text{ where }A(-2,0),B(2,0),C(0,2)\text{ and}
\displaystyle \triangle PQR\text{ where }P(-4,0),Q(4,0),R(0,4)\text{ are similar triangles.}
\displaystyle \hspace{0.2cm}\text{[CBSE 2017]}
\displaystyle \text{Answer:}
\displaystyle \text{To prove the similarity of the two triangles, we need to find the lengths of their }
\displaystyle \text{sides by distance formula and then find the ratio of the corresponding sides.}\displaystyle \text{Distance between any two points }(x_1,y_1)\text{ and }(x_2,y_2)\text{ is}
\displaystyle d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}
\displaystyle \text{In triangle }ABC,
\displaystyle AB=\sqrt{(2-(-2))^2+(0-0)^2}
\displaystyle =\sqrt{16+0}=\sqrt{16}=4
\displaystyle AC=\sqrt{(0-(-2))^2+(2-0)^2}
\displaystyle =\sqrt{4+4}=\sqrt{8}=2\sqrt{2}
\displaystyle BC=\sqrt{(0-2)^2+(2-0)^2}
\displaystyle =\sqrt{4+4}=\sqrt{8}=2\sqrt{2}
\displaystyle \text{In }\triangle PQR,
\displaystyle PQ=\sqrt{(4-(-4))^2+(0-0)^2}
\displaystyle =\sqrt{64+0}=\sqrt{64}=8
\displaystyle PR=\sqrt{(0-(-4))^2+(4-0)^2}
\displaystyle =\sqrt{16+16}=\sqrt{32}=4\sqrt{2}
\displaystyle QR=\sqrt{(0-4)^2+(4-0)^2}
\displaystyle =\sqrt{16+16}=\sqrt{32}=4\sqrt{2}
\displaystyle \text{Now,}
\displaystyle \frac{AB}{PQ}=\frac{4}{8}=\frac{1}{2}
\displaystyle \frac{AC}{PR}=\frac{2\sqrt{2}}{4\sqrt{2}}=\frac{1}{2}
\displaystyle \frac{BC}{QR}=\frac{2\sqrt{2}}{4\sqrt{2}}=\frac{1}{2}
\displaystyle \therefore \frac{AB}{PQ}=\frac{AC}{PR}=\frac{BC}{QR}
\displaystyle \Rightarrow \triangle ABC\sim\triangle PQR
\displaystyle \text{Hence, proved.}
\\

\displaystyle \textbf{Question 38. }\text{In the given figure, }\triangle ABC\text{ is an equilateral triangle of side }3
\displaystyle \text{units. Find the coordinates of the other two vertices.} \hspace{0.2cm}\text{[CBSE 2017]}
\displaystyle \text{Answer:}
\displaystyle  \text{Coordinates of }B\text{ are }(5,0)\text{ as side of equilateral }\triangle\text{ is }3\text{ units.}
\displaystyle \text{Let coordinates of }C\text{ be }(x,y).
\displaystyle \text{Now, }AC^{2}=BC^{2}\qquad [\because AC=BC]
\displaystyle \Rightarrow (x-2)^{2}+(y-0)^{2}=(x-5)^{2}+(y-0)^{2}
\displaystyle \Rightarrow x^{2}+4-4x+y^{2}=x^{2}+25-10x+y^{2}
\displaystyle \Rightarrow 6x=21
\displaystyle \Rightarrow x=\frac{7}{2}
\displaystyle \text{Now, }AC=3\text{ units}
\displaystyle \Rightarrow \sqrt{(x-2)^{2}+(y-0)^{2}}=3
\displaystyle \Rightarrow (x-2)^{2}+(y-0)^{2}=9
\displaystyle \Rightarrow \left(\frac{7}{2}-2\right)^{2}+y^{2}=9
\displaystyle \Rightarrow y^{2}=9-\frac{9}{4}
\displaystyle \Rightarrow y^{2}=\frac{27}{4}\Rightarrow y=\pm\frac{3\sqrt{3}}{2}
\displaystyle \Rightarrow y=\frac{3\sqrt{3}}{2}\quad (+\text{ve sign to be taken})
\displaystyle \therefore \text{Coordinates of }C\text{ are }\left(\frac{7}{2},\frac{3\sqrt{3}}{2}\right).
\\

\displaystyle \textbf{Question 39. }\text{Find the coordinates of the points of trisection of the line segment joining}
\displaystyle \text{the points }(3,-2)\text{ and }(-3,-4). \hspace{0.2cm}\text{[CBSE 2017]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }A(3,-2)\text{ and }D(-3,-4)\text{ be the given points. Let}
\displaystyle B(x_1,y_1)\text{ and }C(x_2,y_2)\text{ be the points of trisection.}
\displaystyle \text{Now, }B\text{ divides }AD\text{ in the ratio of }1:2.\text{ Then,}
\displaystyle x_1=\frac{1\times(-3)+2\times3}{1+2}=1
\displaystyle y_1=\frac{1\times(-4)+2\times(-2)}{1+2}=-\frac{8}{3}
\displaystyle \text{Also, }C\text{ divides }AD\text{ in the ratio of }2:1.\text{ Then,}
\displaystyle x_2=\frac{2\times(-3)+1\times3}{2+1}=-1
\displaystyle y_2=\frac{2\times(-4)+1\times(-2)}{2+1}=-\frac{10}{3}
\displaystyle \text{Hence coordinates of required points are }\left(1,-\frac{8}{3}\right)\text{ and }\left(-1,-\frac{10}{3}\right).
\\

\displaystyle \textbf{Question 40. }\text{In what ratio does the point }P(-4,6)\text{ divide the line segment joining}
\displaystyle \text{the points }A(-6,10)\text{ and }B(3,-8)? \hspace{0.2cm}\text{[CBSE 2017(C)]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }P(-4,6)\text{ divide the line segment }AB\text{ in the ratio }k:1.
\displaystyle \text{Here }A(-6,10)\text{ and }B(3,-8).
\displaystyle \text{Using section formula for x-coordinate,}
\displaystyle -4=\frac{3k-6}{k+1}
\displaystyle -4k-4=3k-6
\displaystyle 7k=2
\displaystyle k=\frac{2}{7}
\displaystyle \text{Also, using y-coordinate,}
\displaystyle 6=\frac{-8k+10}{k+1}
\displaystyle 6k+6=-8k+10
\displaystyle 14k=4
\displaystyle k=\frac{2}{7}
\displaystyle \therefore P(-4,6)\text{ divides }AB\text{ in the ratio }\frac{2}{7}:1=2:7.
\\

\displaystyle \textbf{Question 41. }\text{A line intersects the }y\text{-axis and }x\text{-axis at the points }P\text{ and }Q
\displaystyle \text{respectively. If }(2,-5)\text{ is the mid-point of }PQ,\text{ then find the coordinates of }P\text{ and }Q.
\displaystyle \hspace{0.2cm}\text{[CBSE 2017]}
\displaystyle \text{Answer:}
\displaystyle \text{Since the given line intersects the y-axis at the point }P,\text{ where the abscissa is zero.}
\displaystyle \therefore \text{Coordinates of }P\text{ are }P(0,y).
\displaystyle \text{Also, the same line intersects the x-axis at the point }Q\text{ where the ordinate is zero.}
\displaystyle \therefore \text{Coordinates of }Q\text{ are }Q(x,0).
\displaystyle \text{Let }R(2,-5)\text{ be the mid-point of the line segment }PQ.
\displaystyle \text{By mid-point formula, we have}
\displaystyle 2=\frac{0+x}{2},\qquad -5=\frac{y+0}{2}
\displaystyle \Rightarrow x=4,\ y=-10
\displaystyle \therefore \text{Coordinates of }P\text{ and }Q\text{ are }(0,-10)\text{ and }(4,0)\text{ respectively.}
\\

\displaystyle \textbf{Question 42. }\text{The }x\text{-coordinate of a point }P\text{ is twice its }y\text{-coordinate. If }P
\displaystyle \text{is equidistant from }Q(2,-5)\text{ and }R(-3,6),\text{ find the coordinates of }P.
\displaystyle \hspace{0.2cm}\text{[CBSE 2016]}
\displaystyle \text{Answer:}
\displaystyle \text{Let the required point be }P(2y,y),\ Q(2,-5)\text{ and }R(-3,6)\text{ are given points.}\displaystyle \text{Now,}
\displaystyle PQ=PR
\displaystyle \sqrt{(2y-2)^2+(y+5)^2}=\sqrt{(2y+3)^2+(y-6)^2}\qquad [\because\ \text{using Distance formula}]
\displaystyle \text{Squaring both sides, we get}
\displaystyle 4y^2+4-8y+y^2+10y+25=4y^2+9+12y+y^2-12y+36
\displaystyle \Rightarrow 2y+29=45
\displaystyle \Rightarrow 2y=45-29=16
\displaystyle \Rightarrow y=8
\displaystyle \therefore \text{Coordinates of }P\text{ are }(16,8).
\\

\displaystyle \textbf{Question 43. }\text{Prove that the points }(3,0),(6,4)\text{ and }(-1,3)\text{ are the vertices of a}
\displaystyle \text{right angled isosceles triangle.} \hspace{0.2cm}\text{[CBSE 2016]}
\displaystyle \text{Answer:}
\displaystyle \text{Let the triangle be }\triangle ABC\text{ as shown in figure.}\displaystyle \text{Using distance formula,}
\displaystyle AB=\sqrt{(3-6)^2+(0-4)^2}=5
\displaystyle BC=\sqrt{(-1+6)^2+(4-3)^2}=5\sqrt{2}
\displaystyle CA=\sqrt{(-1-3)^2+(3-0)^2}=5
\displaystyle \text{Here, }AB=AC
\displaystyle \therefore \triangle ABC\text{ is an isosceles triangle.}
\displaystyle \text{Consider, }AB^2+AC^2=(5)^2+(5)^2
\displaystyle =25+25=50
\displaystyle \text{and, }BC^2=(5\sqrt{2})^2=50
\displaystyle \therefore AB^2+AC^2=BC^2
\displaystyle \therefore \triangle ABC\text{ is a right angled triangle in which }\angle A=90^\circ
\displaystyle [\text{By converse of Pythagoras Theorem}]
\\

\displaystyle \textbf{Question 44. }\text{Find the ratio in which }y\text{-axis divides the line segment joining the points}
\displaystyle A(5,-6)\text{ and }B(-1,-4). \text{Also find the coordinates of the point of division.}
\displaystyle \hspace{0.2cm}\text{[CBSE 2016]}
\displaystyle \text{Answer:}
\displaystyle \text{Let the point on y-axis be }P(0,y)\text{ and }AP:PB=k:1.
\displaystyle \text{Coordinates of }P\text{ given by section formula:}
\displaystyle \text{x-coordinate of }P=\frac{5\times 1+(-1)\times k}{k+1}
\displaystyle 0=\frac{5-k}{k+1}
\displaystyle \Rightarrow k=5
\displaystyle \therefore \text{Required ratio is }5:1.
\displaystyle \text{Now, y-coordinate of }P= \frac{(-6)\times 1+(-4)\times k}{k+1}
\displaystyle =\frac{-6-20}{6}=\frac{-13}{3}
\displaystyle \therefore \text{Point on y-axis is }\left(0,-\frac{13}{3}\right).
\\

\displaystyle \textbf{Question 45. }\text{If the point }C(-1,2)\text{ divides internally the line-segment joining}
\displaystyle \text{the points }A(2,5)\text{ and }B(x,y)\text{ in the ratio }3:4,\text{ find the value of }x^{2}+y^{2}.
\displaystyle \hspace{0.2cm}\text{[CBSE 2016]}
\displaystyle \text{Answer:}
\displaystyle \text{Using section formula,}
\displaystyle -1=\frac{3x+4\times2}{3+4}
\displaystyle -1=\frac{3x+8}{7}
\displaystyle -7=3x+8
\displaystyle 3x=-15
\displaystyle x=-5
\displaystyle \text{Similarly,}
\displaystyle 2=\frac{3y+4\times5}{3+4}
\displaystyle 14=3y+20
\displaystyle 3y=-6
\displaystyle y=-2
\displaystyle \therefore x^2+y^2=(-5)^2+(-2)^2=29.

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