The joint equation of a pair of lines passing | vedclass

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(B) Step $1$: Find the equation of the first line. Since it passes through $(1, 4)$ and is parallel to the $X$-axis, its equation is $y = 4$, or $y -

Vedclass · Accepted Answer

$xy - y^2 - 4x + 7y - 12 = 0$

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(B) Step $1$: Find the equation of the first line. Since it passes through $(1, 4)$ and is parallel to the $X$-axis, its equation is $y = 4$, or $y - 4 = 0$.Step $2$: Find the equation of the second line. It passes through $(1, 4)$ and makes an angle of $45^\circ$ with the $X$-axis. Its slope $m = 	an(45^\circ) = 1$. Using the point-slope form $(y - y_1) = m(x - x_1)$, we get $(y - 4) = 1(x - 1)$, which simplifies to $y - 4 = x - 1$, or $x - y + 3 = 0$.Step $3$: The joint equation is the product of the two individual equations: $(y - 4)(x - y + 3) = 0$.Step $4$: Expand the product: $xy - y^2 + 3y - 4x + 4y - 12 = 0$.Step $5$: Simplify to get $xy - y^2 - 4x + 7y - 12 = 0$.

The joint equation of a pair of lines passing through the point $(1, 4)$, where one line is parallel to the $X$-axis and the other makes an angle of $45^\circ$ with the positive direction of the $X$-axis, is:

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