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Joint equation of two lines, through
(2-3), perpendicular to two lines
3x^(2)+xy-2y^(2)=0, is

Joint equation of two lines, through $(2, -3)$ , perpendicular to two lines $3x^{2}+xy-2y^{2}=0$ , is

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Transformations of quadratic functions

Full solution

Q. Joint equation of two lines, through

(2, -3)

, perpendicular to two lines

3x^{2}+xy-2y^{2}=0

, is

Identify slopes by factoring: Identify the slopes of the given lines by factoring the equation $3x^2 + xy - 2y^2 = 0$ . $\newline$ $3x^2 + xy - 2y^2 = (3x + 2y)(x - y) = 0$ $\newline$ So, the slopes of the lines are $-\frac{2}{3}$ and $1$ .
Find perpendicular slopes: Find the slopes of the lines perpendicular to the given lines. $\newline$ The slopes of lines perpendicular to the given lines are the negative reciprocals of the original slopes. $\newline$ So, the slopes of the perpendicular lines are $\frac{3}{2}$ and $-1$ .
Write equations through $(2, -3)$ : Write the equations of the lines passing through $(2, -3)$ with the slopes found in the previous step. $\newline$ For the slope $\frac{3}{2}$ : $y + 3 = \left(\frac{3}{2}\right)(x - 2)$ $\newline$ For the slope $-1$ : $y + 3 = (-1)(x - 2)$
Simplify equations: Simplify the equations of the lines. $\newline$ For the slope $\frac{3}{2}$ : $y = \left(\frac{3}{2}\right)x - 3 - 3$ $\newline$ For the slope $-1$ : $y = -x + 2 - 3$
Correct and simplify equations: Correct the mistake in the previous step and simplify the equations of the lines again. $\newline$ For the slope $\frac{3}{2}$ : $y = \left(\frac{3}{2}\right)x - 3 - 3$ becomes $y = \left(\frac{3}{2}\right)x - 6$ $\newline$ For the slope $-1$ : $y = -x + 2 - 3$ becomes $y = -x - 1$

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