The equation Of the line passing through (1, 5) and making an angle $135^{\circ}$ with the x-axis is

x-y+4=0

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x-y-4=0

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x+y-2=0

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x+y-6=0

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Solution

The correct option is D x+y-6=0
Line passes through the point (1,5).
It makes angle $135^{o}$ with x-axis. Therefore,
Slope, m = $tan 135^{o} = - 1$
Equation of line is
$y - y_{1} = m (x - x_{1})$
$y - 5 = - 1 (x - 1)$
$y - 5 = - x + 1$
$x + y - 6 = 0$
Hence, option D is correct.

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Q. x + y + z = 0
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Q. Solve graphically each of the following systems of linear equations. Also find the coordinates of the points where the lines meet axis of y.

(i) 2x − 5y + 4 = 0,
2x + y − 8 = 0

(ii) 3x + 2y = 12,
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(iii) 2x + y − 11 = 0,
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(iv) x + 2y − 7 = 0,
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If $θ$ is an angle between the lines given by the equation $6 x^{2} + 5 x y - 4 y^{2} + 7 x + 13 y - 3 = 0$ , then equation of the line passing through the point of intersection of these lines and making an angle $θ$ with the positive x - axis is

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The equation Of the line passing through (1, 5) and making an angle 135∘ with the x-axis is

The correct option is D x+y-6=0Line passes through the point (1,5).It makes angle 135o with x-axis. Therefore,Slope, m = tan135o=−1Equation of line isy−y1=m(x−x1)y−5=−1(x−1)y−5=−x+1x+y−6=0Hence, option D is correct.

The equation Of the line passing through (1, 5) and making an angle $135^{\circ}$ with the x-axis is

The correct option is D x+y-6=0
Line passes through the point (1,5).
It makes angle $135^{o}$ with x-axis. Therefore,
Slope, m = $tan 135^{o} = - 1$
Equation of line is
$y - y_{1} = m (x - x_{1})$
$y - 5 = - 1 (x - 1)$
$y - 5 = - x + 1$
$x + y - 6 = 0$
Hence, option D is correct.