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

Answer:

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

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

y-intercept = b

Therefore the equation of line passing through (0,b) and having slope of -\frac{a}{b}  is:

y - b =  -\frac{a}{b} (x-0)

by-b^2 = -ax

ax+by=b^2

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

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

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

Answer:

Mid point of AB =  (\frac{-5+3}{2}, \frac{-3+5}{2}) = (-1, 1)

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

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

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

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

Therefore the equation of altitude of \triangle OAB through B 

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

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

5y+15 = -3x-15

5y+3x+30 = 0

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

Answer:

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

Slope of 3x = y + 1 is 3 

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

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

Mid point of (-2, 3)  and (4, 1) = (\frac{4-2}{2}, \frac{1+3}{2}) = (1,2) 

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

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

Answer:

Given x cos \ 30^o + y sin \ 30^o = 2 

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

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

y-3 =  -\sqrt{3} (x-4) 

y+\sqrt{3} x= 4 \sqrt{3}+3 

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

(i) Perpendicular to the line 2x - y + 7 = 0

(ii) Parallel to it.

Answer:

Given (k-2)x+(k+3)y-5=0

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

Slope of line 2x-y+7 = 0 is 2

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

(i) If perpendicular

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

k+3 = 2k-4

k= 7

(ii) If parallel

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

2k+6 = -k +2 

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

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

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

(ii) The co-ordinates of the point P , where the perpendicular through A , as obtained in (i.), meets BC .

Answer:

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

Slope of line perpendicualr to BC = -\frac{4}{3}

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

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

3y-15 = -4x

3y+4x=15   … … … … (i)

(ii) Equation of BC

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

4y - 28 = 3x - 33

4y-3x= -5    … … … … (ii)

Solving (i) and (ii)  we get x = 3 and $latex  y = 1 &s=0$.

Therefore P is (3, 1)

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l5Question 17: From the given figure, find:

(i) The co-ordinates of A, B, \ and \  C .

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

Answer:

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

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

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

2y-6 = -x+2

2y+x=8

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

Answer:

Mid point of PQ = (\frac{3+7}{2}, \frac{4-2}{2}) = (5,1)

Therefore equation passing through (5,1) and P(-2,-1) is

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

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

7y-7 = 2x-10

7y-2x+3=0

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

Answer:

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

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

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

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

Therefore PB \parallel BC .

PQBC is a trapezoid.

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

(i) The co-ordinates of A \ and \ B .

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

Answer:

AP:PB = 1:2

(i) Let A (0,y) and B(x,0)

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

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

Therefore B(-6, 0) and A (0, -3)

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

Slope of line perpendicular to AB  = 2

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

y - (-2) = 2(x-(-4))

y+2 = 2(x+4)

y = 2x+6

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