Question 1: Point \displaystyle P divides the line segment joining the points \displaystyle A (8, 0) \text{ and } B (16, -8) in the \displaystyle\text{Ratio: } 3 : 5 . Find the co-ordinates of point \displaystyle P . Also, find the equation of the line through \displaystyle P and parallel to \displaystyle 3x + 5y = 7 .

Answer:

\displaystyle\text{Given } P divides \displaystyle A (8, 0) \text{ and } B (16, -8) in the \displaystyle\text{Ratio: } 3 : 5  

\displaystyle\text{Ratio: } m_1:m_2 = 3 : 5  

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

\displaystyle x = \frac{3 \times 16+5 \times 8}{3+5} = \frac{88}{8} = 11  

\displaystyle y = \frac{3 \times (-8)+5 \times 0}{3+5} = \frac{-24}{8} = -3  

\displaystyle\text{Therefore } P = (11, -3)  

\displaystyle\text{Equation of the line given : } 3x+5y=7 \Rightarrow \text{ Slope } = m = \frac{-3}{5}  

Therefore the required equation is

\displaystyle y - y_1 = m(x-x_1)  

\displaystyle y -(-3) = \frac{-3}{5} (x-11) \text{ or } 5y +3x=18  

\displaystyle \\

Question 2: The line segment joining the points \displaystyle A (3, -4) \text{ and } B (-2, 1) is divided in the \displaystyle\text{Ratio: } 1 : 3 at point \displaystyle P in it. Find the co-ordinates of \displaystyle P . Also, find the equation of the line through \displaystyle P and perpendicular to the line \displaystyle 5x - 3y = 4 .

Answer:

\displaystyle\text{Given } P \text{ divides } A (3,-4) \text{ and } B (-2,1) in the \displaystyle\text{Ratio: } 1:3  

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

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

\displaystyle x = \frac{1 \times (-2)+3 \times 3}{1+3} = \frac{7}{4}  

\displaystyle y = \frac{1 \times 1+3 \times (-4)}{1+3} = \frac{-11}{4}  

\displaystyle\text{Therefore } P = ( \frac{7}{4} , \frac{-11}{4} )  

\displaystyle\text{Equation of the line given : } 5x - 3y = 4 \Rightarrow \text{ Slope } = m_1 = \frac{5}{3}  

\displaystyle\text{Therefore the slope of the line perpendicular to the above line } = \frac{-3}{5}  

Therefore the required equation is

\displaystyle y - y_1 = m(x-x_1)  

\displaystyle y - \frac{-11}{4} = \frac{-3}{5} (x- \frac{7}{4} ) \text{ or } 10y +6x+17 = 0  

\displaystyle \\

Question 3: A line \displaystyle 5x + 3y + 15 = 0 meets \displaystyle y-axis at point \displaystyle P . Find the co-ordinates of point \displaystyle P . Find the equation of a line through \displaystyle P and perpendicular to \displaystyle x - 3y + 4 = 0 .

Answer:

At \displaystyle y-axis, x = 0 \Rightarrow y = -5  

Therefore the coordinate of \displaystyle P = (0, -5)  

\displaystyle x - 3y + 4 = 0 \Rightarrow \text{ Slope } = \frac{1}{3}  

Therefore the slope of a like perpendicular to this line \displaystyle = -3  

Hence the line passing through \displaystyle (0, -5) \text{ with a Slope of } -3 \text{ is}  

\displaystyle y-(-5) = -3(x-0) \Rightarrow y+3x+5 = 0  

\displaystyle \\

Question 4: Find the value of \displaystyle k for which the lines \displaystyle kx - 5y + 4 = 0 \text{ and } 5x - 2y + 5 = 0 are perpendicular to each other. [2003]

Answer:

\displaystyle\text{Slope of } kx - 5y + 4 = 0 \Rightarrow m_1 = \frac{k}{5}  

\displaystyle\text{Slope of } 5x - 2y + 5 = 0 \Rightarrow m_2 = \frac{5}{2}  

\displaystyle \text{Since the two lines are perpendicular, } m_1. m_2 = -1  

\displaystyle \Rightarrow \frac{k}{5} . \frac{5}{2} = -1  

\displaystyle \Rightarrow k = -2  

\displaystyle \\

Question 5: A straight line passes through the points \displaystyle P (-1, 4) \text{ and } Q (5, -2) . It intersects the co-ordinate axes at points \displaystyle A \text{ and } B . \displaystyle M is the mid-point of the line segment \displaystyle AB . Find:

The equation of the line

The co-ordinates of \displaystyle A \text{ and } B  

The co-ordinates of \displaystyle M [2003]

Answer:

\displaystyle P (-1, 4) \text{ and } Q (5, -2)  

The equation of the line:

\displaystyle y - 4 = \frac{-2-4}{5-(-1)} (x+1)  

\displaystyle \Rightarrow y-4 = -1 (x+1)  

\displaystyle \Rightarrow y+x=3  

\displaystyle \text{The x-intercept} A = (3,0) \text{ and the y-intercept} B = (0,3)  

\displaystyle \text{The coordinate of } M = ( \frac{3+0}{2} , \frac{0+3}{2} ) = ( \frac{3}{2} , \frac{3}{2} )  

\displaystyle \\

Question 6: \displaystyle (1, 5) \text{ and } (-3, -1) are the co-ordinates of vertices \displaystyle A \text{ and } C respectively of rhombus \displaystyle ABCD . Find the equations of the diagonals \displaystyle AC \text{ and } BD .

Answer:

\displaystyle\text{Midpoint } M = ( \frac{-3+1}{2} , \frac{-1+5}{2} ) = (-1, 2)  

\displaystyle\text{Slope of } AC = \frac{-1-5}{-3-1} = \frac{3}{2}  

Equation of \displaystyle AC :

\displaystyle y - 5 = \frac{3}{2} (x-1) \Rightarrow 3x-2y+7=0  

\displaystyle\text{Slope of } BD = - \frac{2}{3}  

\displaystyle \text{Therefore equation of } BD :

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

\displaystyle \\

Question 7: Show that \displaystyle A (3, 2), B (6, -2) \text{ and } C (2, -5) can be  vertices of a square. Find the coordinates of its fourth vertex \displaystyle D , if \displaystyle ABCD is a square. Without using the coordinates of vertex \displaystyle D , find the equation of side \displaystyle AD of the square and the equation of diagonal \displaystyle BD .

Answer:

\displaystyle\text{Midpoint of } AC = ( \frac{3+2}{2} , \frac{-5+2}{2} ) = ( \frac{5}{2} , \frac{-3}{2} )  

\displaystyle \text{Let the coordinate of }  D \text{ be }  (x, y)  

In a square, the diagonals bisect each other. Therefore

\displaystyle \frac{x+6}{2} = \frac{5}{2} \Rightarrow x = -1  

\displaystyle \frac{y-2}{2} = \frac{-3}{2} \Rightarrow y = -1  

\displaystyle \text{Hence } D \text{ is } (-1, -1)  

\displaystyle \text{ Slope of } AB = \frac{-2-2}{6-3} = \frac{-4}{3}  

\displaystyle \text{Since }  AB \perp AD \text{, Slope of } AD = \frac{3}{4}  

\displaystyle \text{Hence the equation of  }  AD :

\displaystyle y - 2 = \frac{3}{4} (x-3) \Rightarrow 4y-3x+1 = 0  

\displaystyle \text{Slope of } BD = \frac{-2+1}{6+1} = - \frac{1}{7}  

\displaystyle \text{Hence the equation of  }  BD :

\displaystyle y +2 = - \frac{1}{7} (x-6) \Rightarrow 7y+x+8=0  

\displaystyle \\

Question 8: A line through origin meets the line \displaystyle x = 3y + 2 at right angles at point \displaystyle X . find the coordinates of point \displaystyle X .

Answer:

\displaystyle\text{Given } x = 3y + 2 . .. … … … (i)

\displaystyle \text{Slope of line } x = 3y + 2 \text{ is } \frac{1}{3}  

\displaystyle \text{Slope of perpendicular } = -3  

The equation of a line passing through \displaystyle (0,0) and having slope \displaystyle 3 is

\displaystyle y - 0 = -3 (x-0) \Rightarrow y = -3x . .. … … … (i)

Solving equations (i) and (ii)

\displaystyle x= 3 (-3x) + 2 \Rightarrow x = \frac{1}{5}  

\displaystyle \text{Hence } y = - \frac{3}{5}  

\displaystyle \\

Question 9: A straight line passes through the point \displaystyle (3, 2) and the portion of this line, intercepted between the positive axes, is bisected at this point. Find the equation of the line.

Answer:

Let y-intercept be \displaystyle (0, y) and x-intercept be \displaystyle (x, 0)  

\displaystyle\text{Given } (3,2) is the \displaystyle\text{Midpoint of } (0,y) \text{ and } (x,0) . Therefore:

\displaystyle \text{y-intercept} = (0, 4)  

\displaystyle \text{x-intercept} = (6, 0)  

\displaystyle \text{Slope of line } = \frac{0-4}{6-0} = - \frac{2}{3}  

Equation of line:

\displaystyle y-0 = - \frac{2}{3} (x-6) \Rightarrow 3y+2x=12  

\displaystyle \\

Question 10: Find the equation of the line passing through the point of intersection of \displaystyle 7x + 6y = 71 \text{ and } 5x - 8y = -23 ; and perpendicular to the line \displaystyle 4x -2y = 1 .

Answer:

Solve equations

\displaystyle 7x + 6y = 71 . .. … … … (i)

\displaystyle 5x - 8y = -23 . .. … … … (ii)

Multiply (i) by 4 and (ii) by 3 and then add the equations, we get

\displaystyle 43x = 215 \Rightarrow x = 5  

\displaystyle \text{Substituting }  x = 5 in (i) we get

\displaystyle 7(5)+6y=71 \Rightarrow y = 6  

Therefore the intercept is \displaystyle (5, 6)  

\displaystyle \text{Slope of line }  4x -2y = 1 \text{ is } 2  

\displaystyle \text{Therefore the slope of perpendicular } = - \frac{1}{2}  

Hence the equation of the perpendicular:

\displaystyle y - 6 = - \frac{1}{2} (x-5)  

\displaystyle 2y -12 = -x+5  

\displaystyle 2y+x=17  

\displaystyle \\

Question 11: Find the equation of the line which is perpendicular to the line \displaystyle \frac{x}{a} - \frac{y}{b} =1 at the point where this line meets \displaystyle y-axis .

Answer:

\displaystyle \text{Slope of line } \frac{x}{a} - \frac{y}{b} =1 \text{ is } \frac{b}{a}  

\displaystyle \text{Therefore slope of line perpendicular to given line } = - \frac{a}{b}  

\displaystyle \text{y-intercept} = b  

\displaystyle\text{Slope of } - \frac{a}{b} \text{ is }

\displaystyle \text{Therefore the equation of line passing through (0,b) and having } \\ \\ y - b = - \frac{a}{b} (x-0)  

\displaystyle \Rightarrow by-b^2 = -ax  

\displaystyle \Rightarrow ax+by=b^2  

\displaystyle \\

Question 12: \displaystyle O (0, 0), A (3, 5) \text{ and } B (-5, -3) are the vertices of a triangle \displaystyle OAB . Find:

(i) The equation of the median of triangle \displaystyle OAB through vertex \displaystyle O  

(ii) The equation of altitude of triangle \displaystyle OAB through vertex \displaystyle B  

Answer:

\displaystyle\text{Midpoint of } AB = ( \frac{-5+3}{2} , \frac{-3+5}{2} ) = (-1, 1)  

Therefore the equation of median of \displaystyle \triangle OAB through \displaystyle O is

\displaystyle \text{(i) } y - 0 = \frac{1-0}{-1-0} (x-0) \Rightarrow y = -x \text{ or }  y+x = 0  

\displaystyle \text{(ii) } \text{Slope of } OA = \frac{5-0}{3-0} = \frac{5}{3}  

\displaystyle \text{Slope of line perpendicular to } OA = - \frac{3}{5}  

Therefore the equation of altitude of \displaystyle \triangle OAB through \displaystyle B  

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

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

\displaystyle 5y+15 = -3x-15  

\displaystyle 5y+3x+30 = 0  

\displaystyle \\

Question 13: Determine whether the line through points \displaystyle (-2, 3) \text{ and } (4, 1) is perpendicular to the line \displaystyle 3x = y + 1 . Does line \displaystyle 3x = y +1 bisect the line segment joining the two given points?

Answer:

\displaystyle \text{Slope of line passing through } (-2,3) \text{ and } (4,1) = \frac{1-3}{4-(-2)} = - \frac{1}{3}  

\displaystyle\text{Slope of } 3x = y + 1 \text{ is } 3  

\displaystyle \text{Slope of perpendicular } = - \frac{1}{3}  

Therefore line passing through \displaystyle (-2, 3) \text{ and } (4, 1) is perpendicular to \displaystyle 3x = y + 1  

\displaystyle\text{Midpoint of } (-2, 3) \text{ and } (4, 1) = ( \frac{4-2}{2} , \frac{1+3}{2} ) = (1,2)  

\displaystyle \text{Substituting }  (1, 2) in \displaystyle 3x=y+1 we get that it satisfies the equation. \displaystyle\text{Therefore } 3x=y+1 bisects the line joining \displaystyle (-2, 3) \text{ and } (4, 1)  

\displaystyle \\

Question 14: Given a straight line \displaystyle x \cos 30^{\circ} +y \sin 30^{\circ}=2 . Determine the equation of the other line which is parallel to its and passes through \displaystyle (4, 3) .

Answer:

\displaystyle\text{Given } x \cos 30^{\circ} + y \sin 30^{\circ} = 2

\displaystyle \text{Slope of this line } = - \frac{ \cos 30^{\circ}}{ \sin 30^{\circ}} = -\sqrt{3}

Equation of line with slope \displaystyle -\sqrt{3} and passing through \displaystyle (4, 3) is

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

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

\displaystyle \\

Question 15: Find the value of \displaystyle k such that the line \displaystyle (k-2)x+(k+3)y-5=0 is:

(i) Perpendicular to the line \displaystyle 2x - y + 7 = 0 (ii) Parallel to it.

Answer:

\displaystyle\text{Given } (k-2)x+(k+3)y-5=0  

\displaystyle \text{Slope of this line } = - \frac{k-2}{k+3}  

\displaystyle \text{Slope of line } 2x-y+7 = 0 \text{ is } 2  

\displaystyle \text{Slope of line perpendicular to this line } = - \frac{1}{2}  

(i) If perpendicular

\displaystyle - \frac{1}{2} = - \frac{k-2}{k+3}  

\displaystyle k+3 = 2k-4  

\displaystyle k= 7  

(ii) If parallel

\displaystyle 2 = - \frac{k-2}{k+3}  

\displaystyle 2k+6 = -k +2  

\displaystyle 3k = -4 \Rightarrow k = - \frac{4}{3}  

\displaystyle \\

Question 16: The vertices of a triangle \displaystyle ABC are \displaystyle A (0, 5), B (-1, -2) \text{ and } C (11, 7) . Write down the equation of \displaystyle BC . Find:

(i) The equation of the line through \displaystyle A and perpendicular to \displaystyle BC .

(ii) The coordinates of the point \displaystyle P , where the perpendicular through \displaystyle A , as obtained in (i.), meets \displaystyle BC .

Answer:

\displaystyle \text{(i) } \text{Slope of } BC = \frac{7-(-2)}{11-(-1)} = \frac{9}{4} = \frac{3}{4}  

\displaystyle \text{Slope of line perpendicular to } BC = - \frac{4}{3}  

Therefore equation of line passing through \displaystyle A (0,5) with slope \displaystyle - \frac{4}{3} is:

\displaystyle y-5 = - \frac{4}{3} (x-0)  

\displaystyle 3y-15 = -4x  

\displaystyle 3y+4x=15 . .. … … … (i)

(ii) Equation of \displaystyle BC  

\displaystyle y-7 = \frac{3}{4} (x-11)  

\displaystyle 4y - 28 = 3x - 33  

\displaystyle 4y-3x= -5 . .. … … … (ii)

Solving (i) and (ii) we get \displaystyle x = 3 \text{ and } y = 1 .

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

\displaystyle \\

Question 17: From the given figure, find:

(i) The co-ordinates of \displaystyle A, B, \text{ and } C .

(ii) The equation of the line through \displaystyle A and parallel to \displaystyle BC . [2004]

Answer:

\displaystyle\text{Slope of } BC = \frac{0-2}{3-(-1)} = \frac{-2}{4} = - \frac{1}{2}  

The equation of line parallel to \displaystyle BC and passing through \displaystyle A(2,3)  

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

\displaystyle 2y-6 = -x+2  

\displaystyle 2y+x=8  

\displaystyle \\

Question 18: \displaystyle P (3, 4), Q (7, -2) \text{ and } R (-2, -1) are the vertices of triangle \displaystyle PQR . Write down the equation of the median of the triangle through \displaystyle R . [2005]

Answer:

\displaystyle\text{Midpoint of } PQ = ( \frac{3+7}{2} , \frac{4-2}{2} ) = (5,1)  

Therefore equation passing through \displaystyle (5,1) \text{ and } P(-2,-1) is

\displaystyle y-1 = \frac{-1-1}{-2-5} (x-5)  

\displaystyle y-1 = \frac{2}{7} (x-5)  

\displaystyle 7y-7 = 2x-10  

\displaystyle 7y-2x+3=0  

\displaystyle \\

Question 19: \displaystyle A ((8, -6), B (-4, 2) \text{ and } C (0, -10) are vertices of a triangle \displaystyle ABC . If \displaystyle P is the mid-point of \displaystyle AC , use co-ordinate geometry to show that \displaystyle PQ is parallel to \displaystyle BC . Give a special name to quadrilateral \displaystyle PBCQ .

Answer:

\displaystyle \text{Coordinates of } P = ( \frac{-4+8}{2} , \frac{2-6}{2} ) = (2, -2)  

\displaystyle \text{Coordinates of } Q = ( \frac{8+0}{2} , \frac{-6-10}{2} ) = (4,-8)  

\displaystyle\text{Slope of } PQ = \frac{-8-(-2)}{4-2} = \frac{-6}{2} = -3  

\displaystyle\text{Slope of } BC = \frac{-10-2}{0-(-4)} = \frac{-12}{4} = -3  

\displaystyle\text{Therefore } PB \parallel BC .

\displaystyle PQBC is a trapezoid.

\displaystyle \\

Question 20: A line \displaystyle AB meets the \displaystyle x-axis at point \displaystyle A \text{ and } y-axis at point \displaystyle B . The point \displaystyle P (-4, -2) divides the line segment \displaystyle AB internally such that \displaystyle AP : PB = 1 : 2 . Find:

(i) The co-ordinates of \displaystyle A \text{ and } B .

(ii) Equation of the line through \displaystyle P and perpendicular to \displaystyle AB .

Answer:

\displaystyle AP:PB = 1:2  

\displaystyle \text{(i)  Let } A (0,y) \text{ and } B(x,0)  

\displaystyle\text{Therefore } -4 = \frac{1 \times x+2 \times 0}{3} \Rightarrow x = -6  

Similarly, \displaystyle -2 = \frac{1 \times 0+2 \times y}{3} \Rightarrow y= -3  

\displaystyle\text{Therefore } B(-6, 0) \text{ and } A (0, -3)  

\displaystyle \text{(ii) } \text{Slope of } AB = \frac{-3-0}{0-(-6)} = \frac{-3}{6} = \frac{-1}{2}  

Slope of line perpendicular to \displaystyle AB = 2  

Therefore the equation of line passing through \displaystyle P(-4, -2) with slope \displaystyle 2 :

\displaystyle y - (-2) = 2(x-(-4))  

\displaystyle y+2 = 2(x+4)  

\displaystyle y = 2x+6  

\displaystyle \\

Question 21: A line intersects \displaystyle x-axis at point \displaystyle (-2, 0) and cuts off an intercept of \displaystyle 3 units from the positive side of \displaystyle y-axis . Find the equation of the line. [1992]

Answer:

\displaystyle \text{x-intercept} = (-2, 0)  

\displaystyle \text{y-intercept} = (0, 3)  

Equation of line

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

\displaystyle y-3 = \frac{3}{2} x  

\displaystyle 2y -6 = 3x  

\displaystyle 2y = 3x+6  

\displaystyle \\

Question 22: Find the equation of a line passing through the point \displaystyle (2, 3) and having the \displaystyle \text{x-intercept} of \displaystyle 4 units. [2002]

Answer:

\displaystyle \text{x-intercept} = (4, 0)  

Equation of line passing through \displaystyle (2,3) \text{ and } (4, 0)  

\displaystyle y - 0 = \frac{0-3}{4-2} (x-2)  

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

\displaystyle 2y = -3x+12  

\displaystyle 2y + 3x = 12  

\displaystyle \\

Question 23: The given figure (not drawn to scale) shows two straight lines \displaystyle AB \text{ and } CD . If equation of the line \displaystyle AB is y =x+1 and equation of \displaystyle CD \text{ is } y = \sqrt{3}x-1 . Write down the inclination of lines \displaystyle AB \text{ and } CD ; also, find the angle between \displaystyle AB \text{ and } CD [1989]

Answer:

\displaystyle AB: y = x+1  

\displaystyle CD: y = \sqrt{3}x - 1  

\displaystyle\text{Slope of } AB = 1  

\displaystyle \tan \alpha_1 = 1 \Rightarrow \alpha_1 = 45^{\circ}  

\displaystyle\text{Slope of } CD = \sqrt{3}  

\displaystyle \tan \alpha_2 = 1 \Rightarrow \alpha_2 = 60^{\circ}  

\displaystyle\text{Therefore } 45^{\circ}+(180^{\circ}-60^{\circ}) + \theta = 180^{\circ}  

\displaystyle \Rightarrow \theta = 15^{\circ}  

\displaystyle \\

Question 24: Write down the equation of the line whose gradian is \displaystyle \frac{2}{3} and which passes through \displaystyle P , where \displaystyle P divides the line segment joining \displaystyle A (-2, 6) \text{ and } B (3, -4) in the \displaystyle\text{Ratio: } 2 : 3 [2001]

Answer:

\displaystyle\text{Given } P divides the line segment joining \displaystyle A (-2, 6) \text{ and } B (3, -4) in the \displaystyle\text{Ratio: } 2 : 3  

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

Therefore

\displaystyle x = \frac{2 \times 3 + 3 \times (-2)}{2+3} = \frac{6-6}{5} = 0  

\displaystyle y = \frac{2 \times (-4) + 3 \times 6}{2+3} = \frac{-8+18}{5} = 2  

\displaystyle \therefore P(0,2)  

\displaystyle \text{Equation of a line passing through } P(0,2) \text{ with slope } \frac{3}{2}  

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

\displaystyle 2y - 4 = 3x \Rightarrow 2y = 3x+4  

\displaystyle \\

Question 25: The ordinate of a point lying on the line joining points \displaystyle (6, 4) \text{ and } (7, -5) is -23 . Find the co-ordinates of that point.

Answer:

\displaystyle \text{Let the ordinate of a point lying on the line joining points } (6, 4) \text{ and } \\ \\ (7, -5) \text{ is } -23  \text{ be }  (x, -23)  

Equation of line passing through \displaystyle (6,4) \text{ and } (7, -5)  

\displaystyle y - 4 = \frac{-5-4}{7-6} (x-6)  

\displaystyle y-4 = -9(x-6)  

\displaystyle y+9x=58  

Therefore if \displaystyle y = -23 , then

\displaystyle -23+9x=58 \Rightarrow x = 9  

Therefore the point is \displaystyle (9, -23)  

\displaystyle \\

Question 26: Point \displaystyle A \text{ and } B have co-ordinates \displaystyle (7, 3) \text{ and } (1, 9) respectively. Find:

(i) The \displaystyle\text{Slope of } AB  

(ii) The equation of perpendicular bisector of the line segment \displaystyle AB  

(iii) The value of \displaystyle p of (-2, p) lies on it [2008]

Answer:

\displaystyle \text{(i) } \text{Slope of } AB = \frac{9-(-3)}{1-7} = \frac{12}{-6} = -2  

\displaystyle \text{(ii) } \text{Midpoint of } AB = ( \frac{7+1}{2} , \frac{-3+9}{2} ) = (4,3)  

Therefore equation of line passing through \displaystyle (4,3) and slope \displaystyle \frac{1}{2} is

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

\displaystyle 2y -6 = x-4  

\displaystyle 2y= x+2  

\displaystyle \text{(iii) } p \text{ of }  (-2, p) \text{ lies on it }

\displaystyle\text{Therefore } 2(p) = (-2)+2 \Rightarrow p = 0  

\displaystyle \\

Question 27: \displaystyle A \text{ and } B are two points on the \displaystyle x-axis \text{ and } y-axis respectively. \displaystyle P (2, -3) is the mid-point of \displaystyle AB . Find the

(i) Co-ordinates of \displaystyle A \text{ and } B  

\displaystyle \text{(ii) } \text{Slope of line } AB  

(iii) Equation of line \displaystyle AB [2010]

Answer:

\displaystyle \text{Let } A(x,0) \text{ and } B(0, y)  

\displaystyle P(2, -3) is the mid point

\displaystyle \text{(i) } \text{Therefore } 2 = \frac{0+x}{2} \Rightarrow x = 4  

\displaystyle -3 = \frac{y+0}{2} \Rightarrow y = -6  

\displaystyle \text{Hence } A(4,0) \text{ and } B(0, -6)  

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

(iii) Equation of \displaystyle AB  

\displaystyle y = 0 = \frac{3}{2} (x-4)  

\displaystyle 2y = 3x-12  

\displaystyle \\

Question 28: The equation of a line is \displaystyle 3x + 4y - 7 = 0 . Find:

(i) Slope of the line.

(ii) The equation of a line perpendicular to the given line and passing through the intersection of the lines \displaystyle x -y + 2 = 0 \text{ and } 3x + y- 10 = 0 [2010]

Answer:

\displaystyle \text{(i) Slope } = - \frac{3}{4}  

\displaystyle \text{(ii)  Slope of perpendicular } = \frac{4}{3}  

For point of intersection solve \displaystyle x -y + 2 = 0 \text{ and } 3x + y- 10 = 0  

\displaystyle y = 4 \text{ and } x = 2  

\displaystyle \text{Therefore intersection } = (2, 4)  

Therefore equation of line

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

\displaystyle 3y-12 = 4x-8  

\displaystyle 3y = 4x+4  

\displaystyle \\

Question 29: \displaystyle ABCD is a parallelogram where \displaystyle A (x, y), B (5, 8), C (4, 7) \text{ and } D (2, -4) . Find: (i) Co-ordinates of \displaystyle A (ii) Equation of diagonal \displaystyle BD [2011]

Answer:

\displaystyle \text{(i) } \text{Midpoint of } BD = ( \frac{5+2}{2} , \frac{-4+8}{2} )= ( \frac{7}{2} , 2)  

\displaystyle \text{Therefore we have } A(x, 0), O( \frac{7}{2} , 2) \text{ and } C(4, 7)  

\displaystyle O is the \displaystyle\text{Midpoint of } AC as well (diagonals of a parallelogram bisect each other)

\displaystyle \text{Hence } \frac{x+4}{2} = \frac{7}{2} \Rightarrow x = 3  

\displaystyle \text{and } \frac{y+7}{2} = 2 \Rightarrow y = -3  

\displaystyle \text{Hence } A ( 3, -3)  

(ii) Equation of \displaystyle BD  

\displaystyle y - 8 = \frac{-4-8}{2-5} (x-5)  

\displaystyle y-8 = 4(x-5)  

\displaystyle y - 8 = 4x - 20  

\displaystyle y + 12 = 4x  

\displaystyle \\

Question 30: Given equation of line \displaystyle L_1 \text{ is } y = 4 .

(i) Write the \displaystyle \text{Slope of line } L_2 \text{ is } L_2 is the bisector of angle \displaystyle O  

(ii) Write the co-ordinates of point \displaystyle P .

(iii) Find the equation of \displaystyle L_2  

Answer:

\displaystyle L_1 : y = 4  

\displaystyle \text{(i) } \angle \alpha = 45^{\circ}  

Therefore slope \displaystyle m = \tan 45^{\circ} = 1  

\displaystyle \text{(ii) } \text{Therefore } P (4, 4)  

(iii) Equation of line \displaystyle L_2  

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