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

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

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

$\displaystyle \text{Given Slope } m = 3$

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

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

$\displaystyle y-2=3(x-0)$

$\displaystyle \Rightarrow 3x-y+2=0$

$\displaystyle \Rightarrow y = 3x+2$

$\displaystyle \\$

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

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

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

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

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

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

$\displaystyle \Rightarrow y+1=x$

$\displaystyle \Rightarrow x-y=1$

$\displaystyle \\$

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

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

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

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

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

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

$\displaystyle \Rightarrow 4x+3y=0$

$\displaystyle \\$

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 $\displaystyle x-axis.$

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

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

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

$\displaystyle y-4=\sqrt{3}(x-5)$

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

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

$\displaystyle \\$

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

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

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

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

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

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

$\displaystyle \Rightarrow y-1=x$

$\displaystyle \Rightarrow y=x+1$

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

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

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

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

$\displaystyle \Rightarrow y=x-3$

$\displaystyle \\$

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 $\displaystyle x-axis.$

$\displaystyle P \text{ and } Q$ 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 $\displaystyle y-axis.$

$\displaystyle A \text{ and } B$ 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 $\displaystyle AB \text{ and } CD.$

$\displaystyle P (3, 4)$

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

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

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]

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

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 $\displaystyle BC \text{ and } CD.$

$\displaystyle AB$ 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 $\displaystyle x + 2y = 7 \text{ and } x-y=4$

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 $\displaystyle AC.$

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

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 $\displaystyle BC$

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

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

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

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 $\displaystyle (1, 4) \text{ and } (2, 3).$

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

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

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

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:

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

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

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

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 $\displaystyle 6 \text{ and } \text{x-intercept is } 6.$

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

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 $\displaystyle (-3, 2).$

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 $\displaystyle \text{ y-axis }.$

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.

$\displaystyle \text{Given } (-2, 0)$

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

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]

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 $\displaystyle 3:1.$

$\displaystyle m=- \frac{2}{5}$

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

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]

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 $\displaystyle AP : CP = 2 : 3.$

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

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

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:

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

$\displaystyle \Rightarrow y=1$