Question

The line that passes through the points (5, 3)
and (8. A) is perpendicular to the line
6x + 7y + 13 = 0. Determine the value of A.

Answer 1

Answer:

$\frac{13}{2}$

Step-by-step explanation:

slope of the line that passes thru (5, 3) and (8, A)

$= \frac{a - 3}{8 - 5} \\ \frac{a - 3}{3}$

the line that is perpendicular to the given line is

6x + 7y + 13= 0

finding the slop of this line :-

$6x + 7y + 13 = 0 \\ 7y = - 6x - 13 \\ y = - \frac{6}{7} x - \frac{13}{7}$

Comparing with standard equation of a line

y = mx + c

• where m is the slop and c is the y- intercept

m = - 6/ 7

Since, the two lines are perpendicular to each other the product of their slopes will be - 1

$\frac{a - 3}{3} \times \frac{ - 6}{7} \\ \frac{ - 2(a - 3)}{7} = - 1 \\ - 2a + 6 = - 7 \\ - 2a = - 7 - 6 \\ - 2a = - 13 \\ 2a = 13 \\ a = \frac{13}{2}$

( canceling the 2 negative signs on both the sides )