Question

find the equation of the straight line through the A(-1,3)and 1)parallel 2) perpendicular to the straight line through B(2,-5)C(4,-6)​

Answer 1

Answer:

(i) x + 2y - 5 = 0

(ii) 2x - y + 5 = 0

Step-by-step explanation:

Slope = $\frac{y_2-y_1}{x_2-x_1}$

Slope of line joining B and C is:

$\frac{-6-(-5)}{4-2}=\frac{-6+5}{2}=\frac{-1}{2}$

(i): If line through A is parallel to BC, slope of line through A = slope of BC.

Thus, eq.: { $y-y_1=m(x-x_1)$}

= > y - 3 = (-1/2) { x - (-1) }

= > 2(y-3) = -1( x + 1 )

= > 2y - 6 = - x - 1

= > x + 2y - 6 + 1 = 0

= > x + 2y - 5 = 0

(ii): If line through A is perpendicular to BC, slope of line through A = -1/slope of BC. Slope of line though A = -1/(-1/2) = 2

Thus, eq.:

= > ( y - 3 ) = 2( x + 1 )

= > y - 3 = 2x + 2

= > 2x - y + 5 = 0