Question

find the equation of the straight line passing through the point (2, -3) and perpendicular to the line 3x-2y+5=0

Answer 1

GIVEN :–

• A line passing from a point (2, -3) and perpendicular to the line 3x - 2y + 5 = 0.

TO FIND :–

• Equation of line = ?

SOLUTION :–

• Let's find the Slope of given line 3x - 2y + 5 = 0 –

$\\ \implies \sf Slope = \dfrac{ - (coffieciant \: \: of \: \: x)}{(coffieciant \: \: of \: \: y)}$

$\\ \implies \sf n = \dfrac{ - (3)}{( - 2)}$

$\\ \implies \sf n = \dfrac{3}{2}$

• Now Slope of it's perpendicular line –

$\\ \implies \sf m = - \dfrac{1}{n}$

$\\ \implies \sf m = - \dfrac{1}{ \left(\dfrac{3}{2} \right) }$

$\\ \implies \large \boxed{\sf m = - \dfrac{2}{3} }$

• We know that equation of a line which is passing from (a , b) –

$\\ \implies \large \sf(y - b) = m(x - a)$

• Put the values –

$\\ \implies \sf[y - ( - 3)] =- \dfrac{2}{3} (x - 2)$

$\\ \implies \sf y + 3 =- \dfrac{2}{3} (x - 2)$

$\\ \implies \sf 3(y + 3) =- 2(x - 2)$

$\\ \implies \sf 3y + 9=-2x + 4$

$\\ \implies \sf 3y + 2x=4 - 9$

$\\ \implies \sf 2x + 3y = - 5$

$\\ \implies \large \boxed{\sf 2x + 3y + 5 = 0 }$

• Hence , equation of line is 2x + 3y + 5 = 0

Answer 2

Given :-

$\sf{\implies \; A \; straight\; line \; passes \; through\; ( 2, -3) }$

$\sf{\implies It's \; perpendicular \; to \; 3x - 2y + 5 = 0}$

To Find :-

$\sf{\implies \; Equation \; of \; straight\; line }$

Solution :-

For finding equation of any straight line we require

• Point through which line passes .
• The slope of the equation .

So, firstly we have to find out the slope of the given equation .

We will find it by using the equation of other straight line given in the question .

$\sf{\implies 3x - 2y + 5 = 0 }$

$\sf{\implies \; 2y = 3x + 5 }$

$\sf{\implies y = \dfrac{3x}{2} + \dfrac{5}{2} }$

$\sf{\implies \; y = mx + c \; [ where \; m \; is \; slope ]}$

Comparing both equations we get that slope of the second line is

$\sf{\implies \; [slope ] m = \dfrac{3}{2} }$

The relation between slopes of two perpendicular straight lines is

$\sf{\implies \; m_1 . m_2 = -1 }$

We have slope of 2nd line and have to find the slope of 1st line

$\sf{\implies m_1 = \dfrac{1}{m_2}}$

$\sf{\implies \; m_1 =- \dfrac{1}{\dfrac{3}{2} } \rightarrow \; - \dfrac{2}{3} }$

$\sf{\implies \; slope \; of\; line \; is \; -\dfrac{2}{3} }$

Now we have both points as well as slope .

Equation of straight line passing through (2,-3) will be

$\sf{\implies \; [y -y_1] = m[x - x_1] }$

$\sf{\implies \; [y - (-3)] = m[x - 2] }$

$\sf{\implies \; [y +3 ] =- \dfrac{2}{3} [x - 2] }$

$\sf{\implies \; 3[y+3] = -2[x-2] }$

$\sf{\implies \; 3y + 9 = -2x +4 }$

$\sf{\implies \; 2x + 3y = 4 - 9 }$

$\underline{\underline{\sf{\implies \; 2x + 3y + 5 = 0}} }$