Question

Find the equation of the straight line passing through the point (3,-4) and perpendicular to the line 5x+3y-1=0

Answer 1

EXPLANATION.

equation of the straight line passing through

the point (3,-4).

perpendicular to the line = 5x + 3y - 1 = 0.

Equation of slope = y = mx + c.

→ 5x + 3y - 1 = 0

→ 3y = 1 - 5x

→ y = 1/3 - 5x/3.

Slope of perpendicular = -1/m = -1/-5/3 = 3/5.

Formula of perpendicular slope.

→ slope = b/a = 3/5.

Equation of straight line.

$\sf \: : \implies \: (y - y_{1}) = m(x - x_{1})$

Line passes through point (3, -4).

→ ( y - (-4)) = 3/5 ( x - 3).

→ ( y + 4 ) = 3/5 ( x - 3 ).

→ 5 ( y + 4 ) = 3( x - 3 ).

→ 5y + 20 = 3x - 9.

→ 5y - 3x = -29.

Answer 2

Question

Find the equation of the straight line passing through the point ( 3, -4 ) and perpendicular to the line 5x + 3y - 1 = 0

Solution

5x - 3y + 1 = 0

- 3y = -1 - 5x

$\sf \: y = \dfrac { - 1}{ - 3} \dfrac{ - 5x}{ - 3} \\ \\ y = \sf\dfrac{1}{3} + \dfrac{5x}{3} \\ \\ \sf \: y = \frac{5x}{3} + \frac{1}{3} \\ \\ \sf y = mx + c$

$\sf \: so \: slope \: (m) = \dfrac{5}{3}$

$\sf \: for \:a \: perpendicular \: line \: (m) = \dfrac{1}{ - m} = \dfrac{1}{ - 5 /3} \\ \\ = \sf \: \frac{ - 3}{5} = m$

for a line passing through ( 4 , -3 )

$\sf \: y = y_{1} = m(x - x_{1})$

$\sf \: y - ( - 3) = \dfrac{ - 3}{5} \\ \\ \sf \: y + 3 = \frac{ - 3}{5} (x - 4)$

$\sf \: y + 3 = \dfrac{ - 3x}{5} + \dfrac{12}{5} \\ \\ \sf \: y + 3 = \frac{ - 3x + 12}{5} \\ \\ \sf 5y + 15 = - 3x + 12 \\ \\ \sf \: 3x - 12 + 5y + 15 = 0 \\ \\ \sf \: 5y + 3x + 3 = 0$

Therefore, perpendicular line 3x + 5y + 3 = 0