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Equation of a Line Perpendicular to a Given Line

Point A and B...

Question

Point A and B have co-ordinates (7, -3) and (1, 9) respectively. Find :

(i) the slope of AB.

(ii) the equation of perpendicular bisector of the line segment AB. (iii) the value of ‘p’ of (-2, p) lies on it.

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Solution

Given points are A(7, -3) and B(1, 9).

(i) Slope of AB = $fraction numerator y subscript 2 minus y subscript 1 over denominator x subscript 2 minus x subscript 1 end fraction equals fraction numerator 9 plus 3 over denominator 1 minus 7 end fraction equals negative 2$

(ii) Slope of perpendicular bisector = $fraction numerator negative 1 over denominator negative 2 end fraction equals 1 half$

Mid-point of AB = $open parentheses fraction numerator 7 plus 1 over denominator 2 end fraction comma fraction numerator negative 3 plus 9 over denominator 2 end fraction close parentheses$ =(4, 3)

Equation of perpendicular bisector is:

y - 3 = (x - 4)

2y - 6 = x - 4

x - 2y + 2 = 0

(iii) Point (-2, p) lies on x - 2y + 2 = 0.

-2 - 2p + 2 = 0

2p = 0

p = 0

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Point A and B have co-ordinates (7, -3) and (1, 9) respectively. Find : (i) the slope of AB. (ii) the equation of perpendicular bisector of the line segment AB. (iii) the value of ‘p’ of (-2, p) lies on it.

Given points are A(7, -3) and B(1, 9). (i) Slope of AB = (ii) Slope of perpendicular bisector = Mid-point of AB = =(4, 3) Equation of perpendicular bisector is: y - 3 = (x - 4) 2y - 6 = x - 4 x - 2y + 2 = 0 (iii) Point (-2, p) lies on x - 2y + 2 = 0. -2 - 2p + 2 = 0 2p = 0 p = 0

Point A and B have co-ordinates (7, -3) and (1, 9) respectively. Find :

(i) the slope of AB.

(ii) the equation of perpendicular bisector of the line segment AB. (iii) the value of ‘p’ of (-2, p) lies on it.