Class 10: Coordinate Geometry – Very Short Answer Type Questions

$\displaystyle \textbf{Question 1: }\text{Find the distance of a point }P(x,y)\text{ from the origin.}\hfill\text{[CBSE 2018]}$
$\displaystyle \text{Answer:}$
$\displaystyle OP=\sqrt{x^2+y^2}$
$\displaystyle \\$

$\displaystyle \textbf{Question 2: }\text{Write the coordinates of a point }P\text{ on }x\text{-axis which is equidistant}$
$\displaystyle \text{from the points }A(-2,0) \ \text{and }B(6,0).\hfill\text{[CBSE 2019]}$
$\displaystyle \text{Answer:}$
$\displaystyle P=\left(\frac{-2+6}{2},0\right)=(2,0)$
$\displaystyle \therefore P=(2,0).$
$\displaystyle \\$

$\displaystyle \textbf{Question 3: }\text{Find the value(s) of }x,\text{ if the distance between the points }A(0,0)\text{ and }$
$\displaystyle B(x,-4)\text{ is }5\text{ units.} \ \hfill\text{[CBSE 2019]}$
$\displaystyle \text{Answer:}$
$\displaystyle \sqrt{x^2+(-4)^2}=5\Rightarrow x^2+16=25\Rightarrow x^2=9$
$\displaystyle \therefore x=\pm3.$
$\displaystyle \\$

$\displaystyle \textbf{Question 4: }\text{If the distance between the points }(4,k)\text{ and }(1,0)\text{ is }5,$
$\displaystyle \text{ then what can be the possible value of }k? \ \hfill\text{[CBSE 2017]}$
$\displaystyle \text{Answer:}$
$\displaystyle \sqrt{(4-1)^2+(k-0)^2}=5\Rightarrow 9+k^2=25\Rightarrow k^2=16$
$\displaystyle \therefore k=\pm4.$
$\displaystyle \\$

$\displaystyle \textbf{Question 5: }\text{What is the distance between the points }A(\sin\theta-\cos\theta,0)\text{ and }$
$\displaystyle B(0,\sin\theta+\cos\theta)? \ \hfill\text{[CBSE 2015]}$
$\displaystyle \text{Answer:}$
$\displaystyle AB=\sqrt{(0-\sin\theta+\cos\theta)^2+(\sin\theta+\cos\theta)^2}$
$\displaystyle =\sqrt{2(\sin^2\theta+\cos^2\theta)}=\sqrt{2}$
$\displaystyle \therefore AB=\sqrt{2}\text{ units.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 6: }\text{If }A(1,2),\,B(4,3)\text{ and }C(6,6)\text{ are the three vertices of a parallelogram }$
$\displaystyle ABCD,\text{ find the} \ \text{coordinates of fourth vertex }D.\hfill\text{[CBSE 2010]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }D=(x,y).\text{ Since the diagonals of a parallelogram bisect each other,}$
$\displaystyle \left(\frac{1+6}{2},\frac{2+6}{2}\right)=\left(\frac{4+x}{2},\frac{3+y}{2}\right)$
$\displaystyle \therefore x=3,\quad y=5.$
$\displaystyle \therefore D=(3,5).$
$\displaystyle \\$

$\displaystyle \textbf{Question 7: }\text{If }P(2,p)\text{ is the mid-point of the line segment joining the points }A(6,-5)$
$\displaystyle \text{ and }B(-2,11), \ \text{find the value of }p.\hfill\text{[CBSE 2010]}$
$\displaystyle \text{Answer:}$
$\displaystyle p=\frac{-5+11}{2}=3$
$\displaystyle \therefore p=3.$
$\displaystyle \\$

$\displaystyle \textbf{Question 8: }\text{If }P(x,6)\text{ is the mid-point of the line segment joining }A(6,5)\text{ and }$
$\displaystyle B(4,y),\text{ find }y. \ \hfill\text{[CBSE 2010]}$
$\displaystyle \text{Answer:}$
$\displaystyle 6=\frac{5+y}{2}\Rightarrow12=5+y$
$\displaystyle \therefore y=7.$
$\displaystyle \\$

$\displaystyle \textbf{Question 9: }\text{If the distance between the points }(3,0)\text{ and }(0,y)\text{ is }5\text{ units and }$
$\displaystyle y\text{ is positive, then what is the} \ \text{value of }y?\hfill\text{[CBSE 2010]}$
$\displaystyle \text{Answer:}$
$\displaystyle \sqrt{(3-0)^2+(0-y)^2}=5\Rightarrow9+y^2=25\Rightarrow y^2=16$
$\displaystyle \text{Since }y\text{ is positive, }\therefore y=4.$
$\displaystyle \\$

$\displaystyle \textbf{Question 10: }\text{If }P(2,6)\text{ is the mid-point of the line segment joining }A(6,5)\text{ and }$
$\displaystyle B(4,y),\text{ find }y. \ \hfill\text{[CBSE 2010]}$
$\displaystyle \text{Answer:}$
$\displaystyle 6=\frac{5+y}{2}\Rightarrow12=5+y$
$\displaystyle \therefore y=7.$
$\displaystyle \\$

$\displaystyle \textbf{Question 11: }\text{What is the distance between the points }A(c,0)\text{ and }B(0,-c)?\hfill\text{[CBSE 2010]}$
$\displaystyle \text{Answer:}$
$\displaystyle AB=\sqrt{(c-0)^2+\{0-(-c)\}^2}=\sqrt{2c^2}=|c|\sqrt{2}$
$\displaystyle \therefore AB=|c|\sqrt{2}\text{ units.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 12: }\text{Find the value of }a\text{ so that the point }(3,a)\text{ lies on the line represented by}$
$\displaystyle 2x-3y-5=0. \ \hfill\text{[CBSE 2009]}$
$\displaystyle \text{Answer:}$
$\displaystyle 2(3)-3a-5=0\Rightarrow1-3a=0$
$\displaystyle \therefore a=\frac{1}{3}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 13: }\text{Find the distance between the points }\left(-\frac{8}{5},2\right)\text{ and }\left(\frac{2}{5},2\right).\hfill\text{[CBSE 2009]}$
$\displaystyle \text{Answer:}$
$\displaystyle d=\sqrt{\left(\frac{2}{5}+\frac{8}{5}\right)^2+(2-2)^2}=\frac{10}{5}=2$
$\displaystyle \therefore \text{Distance}=2\text{ units.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 14: }\text{Write the ratio in which the line segment joining the points }A(3,-6)\text{ and }$
$\displaystyle B(5,3)\text{ is divided by} \ \text{the }x\text{-axis.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the }x\text{-axis divide }AB\text{ in the ratio }m:n.$
$\displaystyle 0=\frac{3m-6n}{m+n}\Rightarrow3m=6n$
$\displaystyle \therefore m:n=2:1.$
$\displaystyle \\$

$\displaystyle \textbf{Question 15: }\text{Find the values of }x\text{ for which the distance between the points } \\ P(2,-3)\text{ and }Q(x,5)\text{ is }10.$
$\displaystyle \text{Answer:}$
$\displaystyle (x-2)^2+(5+3)^2=10^2$
$\displaystyle (x-2)^2=36\Rightarrow x-2=\pm6$
$\displaystyle \therefore x=8\text{ or }x=-4.$
$\displaystyle \\$

$\displaystyle \textbf{Question 16: }\text{The line segment joining the points }A(4,-5)\text{ and }B(4,5)\text{ is divided by the point }$
$\displaystyle P\text{ such that} \ AP:AB=2:5.\text{ Find the coordinates of }P.\hfill\text{[CBSE 2023]}$
$\displaystyle \text{Answer:}$
$\displaystyle AP:PB=2:3$
$\displaystyle P=\left(\frac{2(4)+3(4)}{5},\frac{2(5)+3(-5)}{5}\right)=(4,-1)$
$\displaystyle \therefore P=(4,-1).$
$\displaystyle \\$

$\displaystyle \textbf{Question 17: }\text{Point }P(x,y)\text{ is equidistant from points }A(5,1)\text{ and }B(1,5).$
$\displaystyle \text{ Prove that }x=y. \ \hfill\text{[CBSE 2023]}$
$\displaystyle \text{Answer:}$
$\displaystyle PA=PB\Rightarrow(x-5)^2+(y-1)^2=(x-1)^2+(y-5)^2$
$\displaystyle -10x-2y=-2x-10y\Rightarrow8y=8x$
$\displaystyle \therefore x=y.$
$\displaystyle \\$

$\displaystyle \textbf{Question 18: }\text{Write the coordinates of a point on }x\text{-axis which is equidistant from the points} \\ (-3,4)\text{ and }(2,5).$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the required point be }P(x,0).\text{ Since }PA=PB,$
$\displaystyle (x+3)^2+4^2=(x-2)^2+5^2$
$\displaystyle 6x+25=-4x+29\Rightarrow10x=4\Rightarrow x=\frac{2}{5}$
$\displaystyle \therefore P=\left(\frac{2}{5},0\right).$
$\displaystyle \\$

$\displaystyle \textbf{Question 19: }\text{Write the coordinates of the point dividing the line segment joining points }$
$\displaystyle (2,3)\text{ and }(3,4) \ \text{internally in the ratio }1:5.$
$\displaystyle \text{Answer:}$
$\displaystyle P=\left(\frac{1(3)+5(2)}{1+5},\frac{1(4)+5(3)}{1+5}\right)$
$\displaystyle \therefore P=\left(\frac{13}{6},\frac{19}{6}\right).$
$\displaystyle \\$

$\displaystyle \textbf{Question 20: }\text{If the distance between points }(x,0)\text{ and }(0,3)\text{ is }5,\text{ what are the values of }x?$
$\displaystyle \text{Answer:}$
$\displaystyle \sqrt{x^2+3^2}=5\Rightarrow x^2=16$
$\displaystyle \therefore x=\pm4.$
$\displaystyle \\$

$\displaystyle \textbf{Question 21: }\text{What is the distance between the points }(5\sin60^\circ,0)\text{ and }(0,5\sin30^\circ)?$
$\displaystyle \text{Answer:}$
$\displaystyle d=\sqrt{\left(\frac{5\sqrt{3}}{2}\right)^2+\left(\frac{5}{2}\right)^2}$
$\displaystyle =\sqrt{\frac{75+25}{4}}=5\text{ units.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 22: }\text{Write the ratio in which the line segment joining points }(2,3)\text{ and } \\ (3,-2)\text{ is divided by }y\text{-axis.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Both points have positive }x\text{-coordinates, so the }y\text{-axis does not intersect the line segment.}$
$\displaystyle \text{If the line is produced, let the }y\text{-axis divide it externally in the ratio }m:n.$
$\displaystyle 0=\frac{3m-2n}{m-n}\Rightarrow3m=2n$
$\displaystyle \therefore m:n=2:3.$
$\displaystyle \\$

$\displaystyle \textbf{Question 23: }\text{Write the distance between the points }A(10\cos\theta,0)\text{ and }B(0,10\sin\theta).$
$\displaystyle \text{Answer:}$
$\displaystyle AB=\sqrt{(10\cos\theta)^2+(10\sin\theta)^2}=10\sqrt{\cos^2\theta+\sin^2\theta}$
$\displaystyle \therefore AB=10\text{ units.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 24: }\text{Find the type of triangle }ABC\text{ formed whose vertices are }A(1,0),\,B(-5,0)$
$\displaystyle \text{ and }C(-2,5). \ \hfill\text{[CBSE 2024]}$
$\displaystyle \text{Answer:}$
$\displaystyle AC=\sqrt{(-2-1)^2+(5-0)^2}=\sqrt{34},\quad BC=\sqrt{(-2+5)^2+(5-0)^2}=\sqrt{34}$
$\displaystyle \therefore AC=BC.$
$\displaystyle \therefore \triangle ABC\text{ is an isosceles triangle.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 25: }\text{In what ratio is the line segment joining the points }(3,-5)\text{ and }(-1,6)$
$\displaystyle \text{ divided by the line }y=x? \ \hfill\text{[CBSE 2024]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the line }y=x\text{ divide the line segment in the ratio }m:n.$
$\displaystyle x=\frac{-m+3n}{m+n},\quad y=\frac{6m-5n}{m+n}$
$\displaystyle \text{Since }y=x,\quad6m-5n=-m+3n\Rightarrow7m=8n$
$\displaystyle \therefore m:n=8:7.$
$\displaystyle \\$

$\displaystyle \textbf{Question 26: }A(3,0),\,B(6,4)\text{ and }C(-1,3)\text{ are vertices of a triangle }ABC.\text{ Find the length of}$
$\displaystyle \text{its median }BE.\hfill\text{[CBSE 2024]}$
$\displaystyle \text{Answer:}$
$\displaystyle E=\left(\frac{3-1}{2},\frac{0+3}{2}\right)=\left(1,\frac{3}{2}\right)$
$\displaystyle BE=\sqrt{(6-1)^2+\left(4-\frac{3}{2}\right)^2}$
$\displaystyle =\sqrt{25+\frac{25}{4}}=\frac{5\sqrt{5}}{2}\text{ units.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 27: }\text{Points }A(-1,y)\text{ and }B(5,7)\text{ lie on a circle with centre }O(2,-3y)\text{ such that}$
$\displaystyle AB\text{ is a diameter of the circle. Find the value of }y.\text{ Also, find the radius of the circle.}\hfill\text{[CBSE 2024]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Since }AB\text{ is a diameter, }O\text{ is the mid-point of }AB.$
$\displaystyle -3y=\frac{y+7}{2}\Rightarrow-6y=y+7\Rightarrow y=-1.$
$\displaystyle \therefore O=(2,3)\text{ and }A=(-1,-1).$
$\displaystyle \text{Radius}=OA=\sqrt{(2+1)^2+(3+1)^2}=\sqrt{25}=5\text{ units.}$
$\displaystyle \therefore y=-1\text{ and the radius is }5\text{ units.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 28: }\text{Write the perimeter of the triangle formed by the points }O(0,0),\,A(a,0) \\ \text{and }B(0,b).$
$\displaystyle \text{Answer:}$
$\displaystyle OA=a,\quad OB=b,\quad AB=\sqrt{a^2+b^2}$
$\displaystyle \therefore \text{Perimeter}=a+b+\sqrt{a^2+b^2}\text{ units.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 29: }\text{If }A(-1,3),\,B(1,-1)\text{ and }C(5,1)\text{ are the vertices of a triangle }ABC,$
$\displaystyle \text{ what is the length of the} \ \text{median through vertex }A?$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Mid-point of }BC=\left(\frac{1+5}{2},\frac{-1+1}{2}\right)=(3,0)$
$\displaystyle \text{Length of median}=\sqrt{(3+1)^2+(0-3)^2}=\sqrt{16+9}=5\text{ units.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 30: }\text{If the mid-point of the segment joining }A(x,y+1)\text{ and }$
$\displaystyle B(x+1,y+2)\text{ is }C\left(\frac{3}{2},\frac{5}{2}\right), \ \text{find }x,y.$
$\displaystyle \text{Answer:}$
$\displaystyle \frac{x+x+1}{2}=\frac{3}{2}\Rightarrow2x+1=3\Rightarrow x=1$
$\displaystyle \frac{y+1+y+2}{2}=\frac{5}{2}\Rightarrow2y+3=5\Rightarrow y=1$
$\displaystyle \therefore x=1,\quad y=1.$
$\displaystyle \\$

$\displaystyle \textbf{Question 31: }\text{Two vertices of a triangle have coordinates }(-8,7)\text{ and }(9,4).\text{ If the}$
$\displaystyle \text{centroid of the triangle is at the origin, what are the coordinates of the third vertex?}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the third vertex be }(x,y).$
$\displaystyle \frac{-8+9+x}{3}=0\Rightarrow x=-1$
$\displaystyle \frac{7+4+y}{3}=0\Rightarrow y=-11$
$\displaystyle \therefore \text{The third vertex is }(-1,-11).$
$\displaystyle \\$

$\displaystyle \textbf{Question 32: }\text{Write the coordinates of the reflections of point }(3,5)\text{ in }x\text{ and }y\text{-axes.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Reflection in }x\text{-axis}=(3,-5)$
$\displaystyle \text{Reflection in }y\text{-axis}=(-3,5).$
$\displaystyle \\$

$\displaystyle \textbf{Question 33: }\text{A line intersects }y\text{-axis and }x\text{-axis at points }P\text{ and }Q\text{ respectively. If }$
$\displaystyle R(2,5)\text{ is the mid-point} \ \text{of line segment }PQ,\text{ then find the coordinates of }P\text{ and }Q.\hfill\text{[CBSE 2023]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }P=(0,y)\text{ and }Q=(x,0).$
$\displaystyle \left(\frac{x}{2},\frac{y}{2}\right)=(2,5)$
$\displaystyle \therefore x=4,\quad y=10.$
$\displaystyle \therefore P=(0,10),\quad Q=(4,0).$
$\displaystyle \\$

$\displaystyle \textbf{Question 34: }\text{Find the coordinates of the point which is equidistant from the three vertices }$
$\displaystyle A(2x,0),\,O(0,0) \ \text{and }B(0,2y)\text{ of }\triangle AOB.$
$\displaystyle \text{Answer:}$
$\displaystyle \triangle AOB\text{ is right-angled at }O.$
$\displaystyle \text{The point equidistant from all three vertices is the mid-point of hypotenuse }AB.$
$\displaystyle \therefore P=\left(\frac{2x+0}{2},\frac{0+2y}{2}\right)=(x,y).$
$\displaystyle \therefore \text{The required point is }(x,y).$
$\displaystyle \\$

$\displaystyle \textbf{Question 35: }\text{If the centroid of the triangle formed by points }P(a,b),\,Q(b,c)\text{ and }R(c,a)$
$\displaystyle \text{ is at the origin, what} \ \text{is the value of }a+b+c?$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Centroid}=\left(\frac{a+b+c}{3},\frac{a+b+c}{3}\right)=(0,0)$
$\displaystyle \therefore a+b+c=0.$
$\displaystyle \\$

$\displaystyle \textbf{Question 36: }\text{In Q. No. 35, what is the value of }\frac{a^2}{bc}+\frac{b^2}{ca}+\frac{c^2}{ab}?$
$\displaystyle \text{Answer:}$
$\displaystyle \frac{a^2}{bc}+\frac{b^2}{ca}+\frac{c^2}{ab}=\frac{a^3+b^3+c^3}{abc}$
$\displaystyle \text{Since }a+b+c=0,\quad a^3+b^3+c^3=3abc.$
$\displaystyle \therefore \frac{a^2}{bc}+\frac{b^2}{ca}+\frac{c^2}{ab}=3.$
$\displaystyle \\$

$\displaystyle \textbf{Question 37: }\text{If points }Q\text{ and }R\text{ are reflections of point }P(-3,4)\text{ in }x \\ \text{and }y\text{-axes respectively, what is }QR?$
$\displaystyle \text{Answer:}$
$\displaystyle Q=(-3,-4),\quad R=(3,4)$
$\displaystyle QR=\sqrt{(3+3)^2+(4+4)^2}=\sqrt{36+64}=10$
$\displaystyle \therefore QR=10\text{ units.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 38: }\text{What are the coordinates of the point where the perpendicular bisector of}$
$\displaystyle \text{the line segment joining the points }A(1,5)\text{ and }B(4,6)\text{ cuts the }y\text{-axis?}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the required point be }P(0,y).\text{ Since }PA=PB,$
$\displaystyle (0-1)^2+(y-5)^2=(0-4)^2+(y-6)^2$
$\displaystyle 1+y^2-10y+25=16+y^2-12y+36$
$\displaystyle 2y=26\Rightarrow y=13$
$\displaystyle \therefore P=(0,13).$
$\displaystyle \\$