Question 1: Find the equation of a line whose: \displaystyle \text{ y-intercept } = 2\text{ and slope } = 3

Answer:

\displaystyle \text{ y-intercept } = 2\text{ and slope } = 3  

\displaystyle \text{Since } \text{ y-intercept } =2 \text{, the corresponding point on } y-axis = (0,2)  

\displaystyle \text{Given Slope } m = 3  

\displaystyle \text{Therefore } m = 3, (x_1, y_1)=(0,2)  

\displaystyle \text{Required equation of the line: } (y-y_1)=m(x-x_1)  

\displaystyle y-2=3(x-0)  

\displaystyle \Rightarrow 3x-y+2=0  

\displaystyle \Rightarrow y = 3x+2  

\displaystyle \\

Question 2: Find the equation of a line whose: \displaystyle \text{ y-intercept } = -1 \text{ and inclination } = 45^{\circ}  

Answer:

\displaystyle \text{ y-intercept } = -1\text{ and slope } = \tan 45^{\circ} = 1  

\displaystyle \text{Since } \text{ y-intercept } =-1 \text{, the corresponding point on y-axis } = (0,-1)  

\displaystyle \text{Therefore } m = 1, (x_1, y_1)=(0,-1)  

\displaystyle \text{Required equation of the line: } (y-y_1)=m(x-x_1)  

\displaystyle y-(-1)=1(x-0)  

\displaystyle \Rightarrow y+1=x  

\displaystyle \Rightarrow x-y=1  

\displaystyle \\

Question 3: Find the equation of the line whose slope is \displaystyle - \frac{4}{3} and which passes through \displaystyle (-3, 4).

Answer:

\displaystyle \text{Slope } = m =- \frac{4}{3}  

\displaystyle \text{Therefore } m =- \frac{4}{3} , (x_1, y_1)=(-3,4)  

\displaystyle \text{Required equation of the line: } (y-y_1)=m(x-x_1)  

\displaystyle y-4=- \frac{4}{3} (x+3)  

\displaystyle \Rightarrow 3y-12=-4x-12  

\displaystyle \Rightarrow 4x+3y=0  

\displaystyle \\

Question 4: Find the equation of the line passing through \displaystyle (5, 4) and makes an angle of \displaystyle 60^{\circ} with the positive direction of \displaystyle x-axis.

Answer:

\displaystyle text{Slope } = m =\tan 60^{\circ} = \sqrt{3}  

\displaystyle \text{Therefore } m =\sqrt{3}, (x_1, y_1)=(5,4)  

\displaystyle \text{Required equation of the line: } (y-y_1)=m(x-x_1)  

\displaystyle y-4=\sqrt{3}(x-5)  

\displaystyle \Rightarrow \sqrt{3}x-y=5\sqrt{3}-4  

\displaystyle \Rightarrow y=\sqrt{3}x+4-5\sqrt{3}  

\displaystyle \\

Question 5: Find the equation of the line passing through:

\displaystyle \text{i) } (0, 1) \text{ and } (1, 2) \hspace{1.0cm} \text{ii) } (-1, -4) \text{ and } (3, 0)  

Answer:

\displaystyle \text{i) } (0, 1) \text{ and } (1, 2)  

\displaystyle \text{Slope } = m = \frac{2-1}{1-0} = \frac{1}{1} = 1  

\displaystyle \text{Required equation of the line: } (y-y_1)=m(x-x_1)  

\displaystyle \Rightarrow y-1=1(x-0)  

\displaystyle \Rightarrow y-1=x  

\displaystyle \Rightarrow y=x+1  

\displaystyle \text{ii) } (-1, -4) \text{ and } (3, 0)  

\displaystyle \text{Slope } = m = \frac{0-(4)}{3-(-1)} = \frac{4}{4} = 1  

\displaystyle \text{Required equation of the line: } (y-y_1)=m(x-x_1)  

\displaystyle \Rightarrow y-0=1(x-3)  

\displaystyle \Rightarrow y=x-3  

\displaystyle \\

Question 6: The co-ordinates of two points \displaystyle P \text{ and } Q are \displaystyle (2, 6) \text{ and } (-3, 5) respectively. Find:

i) The gradient of \displaystyle PQ ;

ii) The equation of \displaystyle PQ  

iii) The co-ordinates of the point where \displaystyle PQ intersects the \displaystyle x-axis.

Answer:

\displaystyle P \text{ and } Q are \displaystyle (2, 6) \text{ and } (-3, 5) respectively

\displaystyle \text{Slope of PQ } = m = \frac{5-6}{-3-2} = \frac{-1}{-5} = \frac{1}{5}  

\displaystyle \text{Required equation of the line: } (y-y_1)=m(x-x_1)  

\displaystyle y-5= \frac{1}{5} (x+3)  

\displaystyle \Rightarrow 5y-25=x+3  

\displaystyle \Rightarrow 5y-3=28  

\displaystyle \\

Question 7: The co-ordinates of two points \displaystyle A \text{ and } B are \displaystyle (-3, 4) \text{ and } (2, -1) . Find:

i) The equation of \displaystyle AB ;

ii) The co-ordinates of the point where the line \displaystyle AB intersect the \displaystyle y-axis.

Answer:

\displaystyle A \text{ and } B are \displaystyle (-3, 4) \text{ and } (2, -1)  

\displaystyle \text{Slope of AB } = m = \frac{-1-4}{2-(-3)} = \frac{-5}{5} = -1  

\displaystyle \text{Required equation of the line: } (y-y_1)=m(x-x_1)  

\displaystyle y-(-1)= -1(x-2)  

\displaystyle \Rightarrow y+1=-x+2  

\displaystyle \Rightarrow y+x=1  

\displaystyle \text{x-intercept }  \Rightarrow y =0  

Therefore when \displaystyle y =0, x = 1  

Hence \displaystyle \text{x-intercept }  = (1,0)  

\displaystyle \\

Question 8: The figure given alongside shows two straight lines \displaystyle AB \text{ and } CD intersecting each other at point \displaystyle P (3, 4) . Find the equations of \displaystyle AB \text{ and } CD.

Answer:

\displaystyle P (3, 4)  

\displaystyle \text{Slope of } AB=m_1= \tan 45^{\circ} = 1  

\displaystyle \text{Slope of } DC=m_2= \tan 60^{\circ} = \sqrt{3}  

Equation of \displaystyle AB:  

\displaystyle y-4=1(x-3)  

\displaystyle \Rightarrow y = x+1  

Equation of \displaystyle DC:  

\displaystyle y-4=\sqrt{3}(x-3)  

\displaystyle \Rightarrow y = \sqrt{3}x+4-3\sqrt{3}  

\displaystyle \\

Question 9: In, \displaystyle A = (3, 5), B = (7, 8) \text{ and } C = (1, -10) . Find the equation of the median through \displaystyle A.  [2013]

Answer:

\displaystyle A = (3, 5), B = (7, 8) \text{ and } C = (1, -10)  

Let \displaystyle D be the mid point of \displaystyle BC . Therefore the coordinates of \displaystyle D are

\displaystyle D =( \frac{1+7}{2} , \frac{-10+8}{2} )=(4, -1)  

\displaystyle \text{Slope of } AD = m = \frac{-1-5}{4-3} = \frac{-6}{1} =-6  

Equation of \displaystyle AD:  

\displaystyle y-(-1)=-6(x-4)  

\displaystyle \Rightarrow y+1=-6x+24  

\displaystyle \Rightarrow y+6x=23  

\displaystyle \\

Question 10: The following figure shows a parallelogram \displaystyle ABCD whose side \displaystyle AB is parallel to the \displaystyle x-axis and vertex \displaystyle C = (7, 5) . Find the equations of \displaystyle BC \text{ and } CD.

Answer:

\displaystyle AB is parallel to \displaystyle x-axis  

\displaystyle \text{Slope of } BC=m_1= \tan 60^{\circ} = \sqrt{3}  

Equation of \displaystyle BC:  

\displaystyle y-5=\sqrt{3}(x-7)  

\displaystyle \Rightarrow y =\sqrt{3}+5-7\sqrt{3}  

\displaystyle \text{Slope of } DC=m_2= \tan 0^{\circ} = 0  

Equation of \displaystyle DC:  

\displaystyle y-5=0(x-7)  

\displaystyle \Rightarrow y=5  

\displaystyle \\

Question 11: Find the equation of the straight line passing through origin and the point of intersection of the lines \displaystyle x + 2y = 7 \text{ and }  x-y=4  

Answer:

Solving \displaystyle x + 2y = 7 \text{ and }  x-y=4 we get \displaystyle x = 5, y = 1  

Hence point of intersection \displaystyle = (5, 1)  

\displaystyle \text{Slope of } m= \tan 0^{\circ} = 0  

\displaystyle \text{Slope of } = m = \frac{1-0}{5-0} = \frac{1}{5}  

\displaystyle y-0= \frac{1}{5} (x-0)  

\displaystyle \Rightarrow x-5y=0  

\displaystyle \\

Question 12: In triangle \displaystyle ABC the co-ordinates of vertices \displaystyle A, B \text{ and } C are \displaystyle (4, 7), (-2, 3) \text{ and } (0, 1) respectively. Find the equation of median through vertex \displaystyle A . Also, find the equation of the line through vertex \displaystyle B and parallel to \displaystyle AC.

Answer:

\displaystyle A = (4, 7), B = (-2, 3) \text{ and } C = (0, 1)  

Let \displaystyle D be the mid point of \displaystyle BC . Therefore the coordinates of \displaystyle D are

\displaystyle D =( \frac{-2+0}{2} , \frac{3+1}{2} )=(-1,2)  

\displaystyle \text{Slope of } AD = m = \frac{2-7}{-1-4} = \frac{-5}{-5} =1  

Equation of \displaystyle AD:  

\displaystyle y-7=1(x-4)  

\displaystyle \Rightarrow y-x=3  

\displaystyle \text{Slope of } AC = m = \frac{7-1}{4-0} = \frac{6}{4} = \frac{3}{2}  

Equation of the line through vertex \displaystyle B and parallel to \displaystyle AC  

\displaystyle y-3= \frac{3}{2} (x+2)  

\displaystyle \Rightarrow 2y-6=3x+6  

\displaystyle \Rightarrow 2y-3x=12  

\displaystyle \\

Question 13: \displaystyle A, B \text{ and } C have co-ordinates \displaystyle (0, 3), (4, 4) \text{ and } (8, 0) respectively. Find the equation of the line through \displaystyle A and perpendicular to \displaystyle BC

Answer:

\displaystyle A(0, 3), B(4, 4) \text{ and } C(8, 0)  

\displaystyle \text{Slope of } BC = m = \frac{0-4}{8-4} = \frac{4}{-4} =-1  

\displaystyle \text{Slope of line perpendicular to } BC = \frac{-1}{-1} = 1  

Therefore the equation of line perpendicular to BC and passing through A:

\displaystyle y-3=1(x-0)  

\displaystyle \Rightarrow y=x+3  

\displaystyle \\

Question 14: Find he equation of the perpendicular dropped from the point \displaystyle (-1, 2) onto the line joining the points \displaystyle (1, 4) \text{ and } (2, 3).

Answer:

\displaystyle A(1, 4), B(2,3) \& C(-1,2)  

\displaystyle \text{Slope of } AB = m = \frac{3-4}{2-1} = \frac{-1}{1} =-1  

\displaystyle \text{Slope of line perpendicular to } BC = \frac{-1}{-1} = 1  

Therefore the equation of line perpendicular to AB and passing through C:

\displaystyle y-2=1(x+1)  

\displaystyle \Rightarrow y=x+3  

\displaystyle \\

Question 15: Find the equation of the line, whose:

\displaystyle \text{i) } \text{ x-intercept } = 5 \text{ and } \text{ y-intercept } = 3  

\displaystyle \text{ii) } \text{ x-intercept } = - 4 \text{ and } \text{ y-intercept } = 6  

\displaystyle \text{iii) } \text{ x-intercept } = - 8 \text{ and } \text{ y-intercept } = - 4  

Answer:

i) Points given are \displaystyle (5,0) \text{ and } (0,3)  

\displaystyle \text{Slope } = \frac{3-0}{0-5} = - \frac{3}{5}  

Equation of line:

\displaystyle y-3=- \frac{3}{5} (x-0)  

\displaystyle \Rightarrow 5y-15=-3x  

\displaystyle \Rightarrow 5y+3x=15  

ii) Points given are \displaystyle (-4,0) \text{ and } (0,6)  

\displaystyle \text{Slope } = \frac{6-0}{0-(-4)} = \frac{3}{2}  

Equation of line:

\displaystyle y-6= \frac{3}{2} (x-0)  

\displaystyle \Rightarrow 2y-12=3x  

\displaystyle \Rightarrow 2y=3x+12  

iii) Points given are \displaystyle (-8,0) \text{ and } (0,-4)  

\displaystyle \text{Slope } m= \frac{-4-0}{0-(-8)} = \frac{-4}{8} = \frac{-1}{2}  

Equation of line:

\displaystyle y-(-4)=- \frac{1}{2} (x-0)  

\displaystyle \Rightarrow 2y+8=-x  

\displaystyle \Rightarrow x+2y+8=0  

\displaystyle \\

Question 16: Find the equation of line whose slope is \displaystyle 6 \text{ and }  \text{x-intercept is } 6.

Answer:

\displaystyle \text{Slope } = - \frac{5}{6} \text{, Intercept} = (6,0)  

Equation of the line:

\displaystyle y-0=- \frac{5}{6} (x-6)  

\displaystyle \Rightarrow 6y=-5x+30  

\displaystyle \Rightarrow 6y+5x=30  

\displaystyle \\

Question 17: Find the equation of the line with \displaystyle \text{ x-intercept } 5 and a point on it \displaystyle (-3, 2).  

Answer:

Given point are \displaystyle (5,0) \text{ and } (-3,2)  

\displaystyle \text{Slope } m= \frac{2-0}{-3-5} = \frac{-2}{8} = \frac{-1}{4}  

Equation of the line:

\displaystyle y-2= \frac{-1}{4} (x+3)  

\displaystyle \Rightarrow 4y-8=-x-3  

\displaystyle \Rightarrow 4y+x=5  

\displaystyle \\

Question 18: Find the equation of the line through \displaystyle (1, 3) and making an intercept of \displaystyle 5 on the \displaystyle \text{ y-axis }.

Answer:

Given point are \displaystyle (1,3) \text{ and } (0,5)  

\displaystyle \text{Slope } m= \frac{5-3}{0-1} = \frac{2}{-1} = -2  

Equation of the line:

\displaystyle y-5=-2(x-0)  

\displaystyle \Rightarrow y+2x=5  

\displaystyle \\

Question 19: Find the equations of the lines passing through the point \displaystyle (-2, 0) and equally inclined to the co-ordinate axes.

Answer:

\displaystyle \text{Given } (-2, 0)  

\displaystyle \text{Slope } = m = \tan 45^{\circ} = 1  

Equation of line:

\displaystyle y-0=1(x+2)  

\displaystyle \Rightarrow y = x+2  

\displaystyle \text{Also Slope } = m = \tan 135^{\circ} = -1  

Equation of line:

\displaystyle y-0=-1(x+2)  

\displaystyle \Rightarrow y + x+2= 0  

\displaystyle \\

Question 20: The line through \displaystyle P (5, 3) intersects \displaystyle \text{ y-axis at }  Q .

i) Write the slope of the line.

ii) Write the equation of the line.

iii) Find the co-ordinates of \displaystyle Q [2012]

Answer:

Given points \displaystyle P(-2, 0) \text{ and } Q(0, y)  

\displaystyle \text{Slope } = m = \tan 45^{\circ} = 1  

Equation of line:

\displaystyle y-3=1(x-5)  

\displaystyle \Rightarrow y = x-2  

When \displaystyle x=0, y = -2  

Hence the co-ordinates of \displaystyle Q = (0, -2)  

\displaystyle \\

Question 21: Write down the equation of the line whose gradient is -and which passes through point \displaystyle P where \displaystyle P divides the line segment joining \displaystyle A (4, -8) \text{ and } B (12, 0) in the ratio \displaystyle 3:1.  

Answer:

\displaystyle m=- \frac{2}{5}  

\displaystyle \text{Ratio: } m_1:m_2 = 3:1  

Let the coordinates of the point \displaystyle P \text{ be } (x, y)  

Therefore

\displaystyle x = \frac{3 \times 12+1 \times 4}{1+2} =10  

\displaystyle y = \frac{3\times 0+1 \times (-8)}{1+2} = -2  

\displaystyle \text{Therefore } P = (10, -2)  

Equation of line:

\displaystyle y-(-2)=- \frac{2}{5} (x-10)  

\displaystyle \Rightarrow 5y+2x=10  

\displaystyle \\

Question 22: \displaystyle A (1, 4), B (3, 2) \text{ and } C (7, 5) are vertices of a triangle \displaystyle ABC. Find:

i) The co-ordinates of the centroid of a triangle \displaystyle ABC.

ii) The equation of a line through the centroid and parallel to \displaystyle AB. [2002]

Answer:

Let \displaystyle O be the centroid. Therefore the coordinates of \displaystyle O are:

\displaystyle O=( \frac{1+3+7}{3} , \frac{4+2+5}{3} )=( \frac{11}{3} , \frac{11}{3} )  

\displaystyle \text{Slope } m= \frac{2-4}{3-1} = \frac{-2}{2} = -1  

Therefore the equation of a line parallel to \displaystyle AB will pass through \displaystyle ( \frac{11}{3} , \frac{11}{3} )  

Equation of the line:

\displaystyle y- \frac{11}{3} =-1(x- \frac{11}{3} )  

\displaystyle \Rightarrow 3y-11=-(3x-11)  

\displaystyle \Rightarrow 3y+3x=22  

\displaystyle \\

Question 23: \displaystyle A (7, -1), B (4, 1) \text{ and } C (-3, 4) are the vertices of a triangle \displaystyle ABC . Find the equation of a line through the vertex \displaystyle B and the point \displaystyle P in \displaystyle AC ; such that \displaystyle AP : CP = 2 : 3.

Answer:

\displaystyle A(7, -1) \text{ and } (-3,4)  

\displaystyle \text{Ratio: } m_1:m_2 = 2:3  

Let the coordinates of the point \displaystyle P be (x, y)  

Therefore

\displaystyle x = \frac{2 \times (-3)+3 \times 7}{2+3} =3  

\displaystyle y = \frac{2\times 4+3 \times (-1)}{2+3} = 1  

\displaystyle \text{Therefore } P = (3, 1)  

\displaystyle \text{Slope } m= \frac{1-1}{3-4} = 0  

Equation of the line:

\displaystyle y-1=0(x-3)  

\displaystyle \Rightarrow y=1