Find the equation of a line passing through t | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Step $1$: Find the point of intersection of $2x - 3y + 4 = 0$ and $3x + 4y - 5 = 0$. Solving these equations,we get $x = -\frac{1}{17}$ and $y = \

Vedclass · Accepted Answer

$119x + 102y = 125$

Vedclass · Answer

(B) Step $1$: Find the point of intersection of $2x - 3y + 4 = 0$ and $3x + 4y - 5 = 0$. Solving these equations,we get $x = -\frac{1}{17}$ and $y = \frac{22}{17}$. The point is $(-\frac{1}{17}, \frac{22}{17})$.Step $2$: The given line is $6x - 7y + 3 = 0$,which has a slope of $m_1 = \frac{6}{7}$.Step $3$: The slope of the line perpendicular to it is $m_2 = -\frac{1}{m_1} = -\frac{7}{6}$.Step $4$: Using the point-slope form $y - y_1 = m(x - x_1)$,we have $y - \frac{22}{17} = -\frac{7}{6}(x + \frac{1}{17})$.Step $5$: Multiplying by $102$,we get $102y - 132 = -119x - 7$,which simplifies to $119x + 102y = 125$.

Find the equation of a line passing through the point of intersection of the lines $2x - 3y + 4 = 0$ and $3x + 4y - 5 = 0$ and perpendicular to the line $6x - 7y + 3 = 0$.

Similar Questions

The lines $x+2ay+a=0$,$x+3by+b=0$,and $x+4cy+c=0$ are concurrent. Then $a, b, c$ are in:

If the lines $x + 2ay + a = 0$,$x + 3by + b = 0$ and $x + 4cy + c = 0$ are concurrent,then $a$,$b$ and $c$ are in

If the lines $4x + 3y - 1 = 0$,$x - y + 5 = 0$,and $kx + 5y - 3 = 0$ are concurrent,then $k=$

Find the equation of the line passing through the point of intersection of the lines $x + y - 2 = 0$ and $2x - y + 1 = 0$ and the origin $(0, 0)$.

If $(x_1, y_1)$ are the roots of $x^2 + 8x - 20 = 0$,$(x_2, y_2)$ are the roots of $4x^2 + 32x - 57 = 0$ and $(x_3, y_3)$ are the roots of $9x^2 + 72x - 112 = 0$,then the points $(x_1, y_1), (x_2, y_2)$ and $(x_3, y_3)$:

Vedclass Test Series

Exam Paper Generator

Online Exam Module