Question

Find the equation of a straight line passing through the point (1,-2) and is
perpendicular to the straight line 2x −5y−7 = 0.

Answer 1

Answer:

Equation of given straight line is:

2x + 5y + 2 = 0

Thus, 5y = -2x -2

Thus, y = (-2/5)*x - (2/5).

Comparing the above form of equation with y = mx + c.

Slope of the given line is = (-2/5)

Slope of the line perpendicular to the above line would be -1/(-2/5) = 5/2

Thus the equation of the line to be found (which would be perpendicular to 2x + 5y + 2 = 0 would be: y = (5/2)*x + c [ Equation of a straight line with slope m is given by y = mx + c where c is the intercept of the line on y axis.]

Since the line passes through (-1,4), substituting x=-1 & y =4,

4 = (5/2)*(-1) + c = c - (5/2)

c = 4 + (5/2) = 13/2

Thus, the equation of line perpendicular to 2x + 5y +2 = 0 which also passes through the point (-1,4) is:

y = (5/2)*x + 13/2

Thus, 2y = 5x + 13

Thus, 5x - 2y + 13=0 is the equation of the line which is perpendicular to the line [2x + 5y + 2=0] & passes thru the point (-1,4).