88672
A square is formed by the lines \(x=0, y=0, x=1\), \(y=1\). Then, the equation of its diagonals will be
1 \(y=x, x+y=2\)
2 \(2 \mathrm{y}=x, x+y=\frac{1}{2}\)
3 \(\mathrm{y}=x, x+y=1\)
4 \(y=2 x, z+y=\frac{1}{4}\)
Explanation:
(C) : Equation of \(\mathrm{OB}\)
\(y=\left(\frac{1-0}{1-0}\right) x\)
\(\Rightarrow \quad(\mathrm{y}=\mathrm{x})\)
Equation of \(\mathrm{AC}\)
\(y-0=-1(x-1)\)
\(\Rightarrow \quad y=-x+1 \Rightarrow x+y=1\)
AP EAMCET-2020-17.09.2020
Straight Line
88673
Find the value of ' \(k\) ' if the angle between the straight lines represented by
\(2 x^{2}+5 x y+3 y^{2}+6 x+7 y+4=0 \text { is } \tan ^{-1}(k) .\)
1 \(\left(\frac{1}{5}\right)\) only
2 \(\left(\frac{-1}{5}\right)\) only
3 \(\pm \frac{1}{5}\)
4 0
Explanation:
(C) : Given,
Straight lines \(2 x^{2}+5 x y+3 y^{2}+6 x+7 y+4=0\)
Angle between straight lines represented by
\(\tan \theta =\frac{3 \sqrt{\mathrm{h}^{2}-\mathrm{ab}}}{\mathrm{a}+\mathrm{b}}\)
\(\tan \theta =\frac{2 \sqrt{(5 / 2)^{2}-6}}{3}= \pm \frac{1}{5}\)
\(a x^{2}+b y^{2}+2 g x+2 f y+c+2 h x y=0 \text { is }\)
Hence,
\(\theta=\tan ^{-1} \mathrm{k}=\tan ^{-1}\left( \pm \frac{1}{5}\right) \Rightarrow\left(\mathrm{k}= \pm \frac{1}{5}\right)\)
AP EAMCET-2020-17.09.2020
Straight Line
88689
A variable line passes through the fixed point \((\alpha, \beta)\) The locus of the foot of the perpendicular from the origin on the line is
1 \(x^{2}+y^{2}-\alpha x-\beta y=0\)
2 \(x^{2}+y^{2}+2 \alpha x+2 \beta y=0\)
3 \(\alpha x+\beta y \pm \sqrt{\left(\alpha^{2}+\beta^{2}\right)}=0\)
(A) : Let \((\alpha, \beta)\) be the given point, let \(Q(x, y)\) be the foot of the perpendicular, and let \(\mathrm{O}\) be the origin.
The line can have any direction
\(\angle \mathrm{PQO}=90^{\circ}\)
Point \(\mathrm{Q}\) lies on the circle having diameter \(\mathrm{OP}\), The locus of point \(\mathrm{Q}\).
\((x-0)(x-\alpha)+(y-0)(y-\beta)=0\)
\(\Rightarrow \quad x^{2}-x \alpha+y^{2}-y \beta=0\)
\(\Rightarrow \quad x^{2}+y^{2}-\alpha x-\beta y=0\)
WB JEE-2019
Straight Line
88647
If the midpoint of the section of the straight line intercepted between the axes is \((1,1)\) then the equation of the line is
1 \(2 x+y=3\)
2 \(2 x-y=1\)
3 \(x-y=0\)
4 \(x+y=2\)
Explanation:
(D) : It is given that \((1,1)\) is the midpoint of line \(\mathrm{AB}\).
\(\Rightarrow 1=\frac{a+0}{2}, 1=\frac{0+\mathrm{b}}{2}\)
\(\Rightarrow a=2 \Rightarrow b=2\)
Equation of line \(\mathrm{AB}\) is
\(\frac{\mathrm{x}}{2}+\frac{\mathrm{y}}{2}=1 \Rightarrow \mathrm{x}+\mathrm{y}=2\)
COMEDK-2011
Straight Line
88648
The perpendicular distance of a line from the origin is \(7 \mathrm{~cm}\) and its slope is -1 . The equation of line is
1 \(x+y+7 \sqrt{2}=0\)
2 \(x-y+7 \sqrt{2}=0\)
3 \(x+y-7 \sqrt{2}=0\)
4 \(x-y-7 \sqrt{2}=0\)
Explanation:
(C) : Equation of line in slope form is\(\mathrm{y}=\mathrm{mx}+\mathrm{c}\)
We have, slope \(=-1\)
So, equation is \(\mathrm{y}=-\mathrm{x}+\mathrm{c}\)
\(\mathrm{x}+\mathrm{y}-\mathrm{c}=0\)
Perpendicular distance of this line from
Origin is \(=7 \mathrm{~cm}\).
Then,
Perpendicular distance from the origin \((0,0)\) to the
Line, \(\quad \mathrm{x}+\mathrm{y}-\mathrm{c}=0\) is-
\(7 =\left|\frac{0+0-c}{\sqrt{2}}\right|\)
\(7 \sqrt{2} =c\)
So, equation are-
\(x-y-7 \sqrt{2}=0\)
88672
A square is formed by the lines \(x=0, y=0, x=1\), \(y=1\). Then, the equation of its diagonals will be
1 \(y=x, x+y=2\)
2 \(2 \mathrm{y}=x, x+y=\frac{1}{2}\)
3 \(\mathrm{y}=x, x+y=1\)
4 \(y=2 x, z+y=\frac{1}{4}\)
Explanation:
(C) : Equation of \(\mathrm{OB}\)
\(y=\left(\frac{1-0}{1-0}\right) x\)
\(\Rightarrow \quad(\mathrm{y}=\mathrm{x})\)
Equation of \(\mathrm{AC}\)
\(y-0=-1(x-1)\)
\(\Rightarrow \quad y=-x+1 \Rightarrow x+y=1\)
AP EAMCET-2020-17.09.2020
Straight Line
88673
Find the value of ' \(k\) ' if the angle between the straight lines represented by
\(2 x^{2}+5 x y+3 y^{2}+6 x+7 y+4=0 \text { is } \tan ^{-1}(k) .\)
1 \(\left(\frac{1}{5}\right)\) only
2 \(\left(\frac{-1}{5}\right)\) only
3 \(\pm \frac{1}{5}\)
4 0
Explanation:
(C) : Given,
Straight lines \(2 x^{2}+5 x y+3 y^{2}+6 x+7 y+4=0\)
Angle between straight lines represented by
\(\tan \theta =\frac{3 \sqrt{\mathrm{h}^{2}-\mathrm{ab}}}{\mathrm{a}+\mathrm{b}}\)
\(\tan \theta =\frac{2 \sqrt{(5 / 2)^{2}-6}}{3}= \pm \frac{1}{5}\)
\(a x^{2}+b y^{2}+2 g x+2 f y+c+2 h x y=0 \text { is }\)
Hence,
\(\theta=\tan ^{-1} \mathrm{k}=\tan ^{-1}\left( \pm \frac{1}{5}\right) \Rightarrow\left(\mathrm{k}= \pm \frac{1}{5}\right)\)
AP EAMCET-2020-17.09.2020
Straight Line
88689
A variable line passes through the fixed point \((\alpha, \beta)\) The locus of the foot of the perpendicular from the origin on the line is
1 \(x^{2}+y^{2}-\alpha x-\beta y=0\)
2 \(x^{2}+y^{2}+2 \alpha x+2 \beta y=0\)
3 \(\alpha x+\beta y \pm \sqrt{\left(\alpha^{2}+\beta^{2}\right)}=0\)
(A) : Let \((\alpha, \beta)\) be the given point, let \(Q(x, y)\) be the foot of the perpendicular, and let \(\mathrm{O}\) be the origin.
The line can have any direction
\(\angle \mathrm{PQO}=90^{\circ}\)
Point \(\mathrm{Q}\) lies on the circle having diameter \(\mathrm{OP}\), The locus of point \(\mathrm{Q}\).
\((x-0)(x-\alpha)+(y-0)(y-\beta)=0\)
\(\Rightarrow \quad x^{2}-x \alpha+y^{2}-y \beta=0\)
\(\Rightarrow \quad x^{2}+y^{2}-\alpha x-\beta y=0\)
WB JEE-2019
Straight Line
88647
If the midpoint of the section of the straight line intercepted between the axes is \((1,1)\) then the equation of the line is
1 \(2 x+y=3\)
2 \(2 x-y=1\)
3 \(x-y=0\)
4 \(x+y=2\)
Explanation:
(D) : It is given that \((1,1)\) is the midpoint of line \(\mathrm{AB}\).
\(\Rightarrow 1=\frac{a+0}{2}, 1=\frac{0+\mathrm{b}}{2}\)
\(\Rightarrow a=2 \Rightarrow b=2\)
Equation of line \(\mathrm{AB}\) is
\(\frac{\mathrm{x}}{2}+\frac{\mathrm{y}}{2}=1 \Rightarrow \mathrm{x}+\mathrm{y}=2\)
COMEDK-2011
Straight Line
88648
The perpendicular distance of a line from the origin is \(7 \mathrm{~cm}\) and its slope is -1 . The equation of line is
1 \(x+y+7 \sqrt{2}=0\)
2 \(x-y+7 \sqrt{2}=0\)
3 \(x+y-7 \sqrt{2}=0\)
4 \(x-y-7 \sqrt{2}=0\)
Explanation:
(C) : Equation of line in slope form is\(\mathrm{y}=\mathrm{mx}+\mathrm{c}\)
We have, slope \(=-1\)
So, equation is \(\mathrm{y}=-\mathrm{x}+\mathrm{c}\)
\(\mathrm{x}+\mathrm{y}-\mathrm{c}=0\)
Perpendicular distance of this line from
Origin is \(=7 \mathrm{~cm}\).
Then,
Perpendicular distance from the origin \((0,0)\) to the
Line, \(\quad \mathrm{x}+\mathrm{y}-\mathrm{c}=0\) is-
\(7 =\left|\frac{0+0-c}{\sqrt{2}}\right|\)
\(7 \sqrt{2} =c\)
So, equation are-
\(x-y-7 \sqrt{2}=0\)
88672
A square is formed by the lines \(x=0, y=0, x=1\), \(y=1\). Then, the equation of its diagonals will be
1 \(y=x, x+y=2\)
2 \(2 \mathrm{y}=x, x+y=\frac{1}{2}\)
3 \(\mathrm{y}=x, x+y=1\)
4 \(y=2 x, z+y=\frac{1}{4}\)
Explanation:
(C) : Equation of \(\mathrm{OB}\)
\(y=\left(\frac{1-0}{1-0}\right) x\)
\(\Rightarrow \quad(\mathrm{y}=\mathrm{x})\)
Equation of \(\mathrm{AC}\)
\(y-0=-1(x-1)\)
\(\Rightarrow \quad y=-x+1 \Rightarrow x+y=1\)
AP EAMCET-2020-17.09.2020
Straight Line
88673
Find the value of ' \(k\) ' if the angle between the straight lines represented by
\(2 x^{2}+5 x y+3 y^{2}+6 x+7 y+4=0 \text { is } \tan ^{-1}(k) .\)
1 \(\left(\frac{1}{5}\right)\) only
2 \(\left(\frac{-1}{5}\right)\) only
3 \(\pm \frac{1}{5}\)
4 0
Explanation:
(C) : Given,
Straight lines \(2 x^{2}+5 x y+3 y^{2}+6 x+7 y+4=0\)
Angle between straight lines represented by
\(\tan \theta =\frac{3 \sqrt{\mathrm{h}^{2}-\mathrm{ab}}}{\mathrm{a}+\mathrm{b}}\)
\(\tan \theta =\frac{2 \sqrt{(5 / 2)^{2}-6}}{3}= \pm \frac{1}{5}\)
\(a x^{2}+b y^{2}+2 g x+2 f y+c+2 h x y=0 \text { is }\)
Hence,
\(\theta=\tan ^{-1} \mathrm{k}=\tan ^{-1}\left( \pm \frac{1}{5}\right) \Rightarrow\left(\mathrm{k}= \pm \frac{1}{5}\right)\)
AP EAMCET-2020-17.09.2020
Straight Line
88689
A variable line passes through the fixed point \((\alpha, \beta)\) The locus of the foot of the perpendicular from the origin on the line is
1 \(x^{2}+y^{2}-\alpha x-\beta y=0\)
2 \(x^{2}+y^{2}+2 \alpha x+2 \beta y=0\)
3 \(\alpha x+\beta y \pm \sqrt{\left(\alpha^{2}+\beta^{2}\right)}=0\)
(A) : Let \((\alpha, \beta)\) be the given point, let \(Q(x, y)\) be the foot of the perpendicular, and let \(\mathrm{O}\) be the origin.
The line can have any direction
\(\angle \mathrm{PQO}=90^{\circ}\)
Point \(\mathrm{Q}\) lies on the circle having diameter \(\mathrm{OP}\), The locus of point \(\mathrm{Q}\).
\((x-0)(x-\alpha)+(y-0)(y-\beta)=0\)
\(\Rightarrow \quad x^{2}-x \alpha+y^{2}-y \beta=0\)
\(\Rightarrow \quad x^{2}+y^{2}-\alpha x-\beta y=0\)
WB JEE-2019
Straight Line
88647
If the midpoint of the section of the straight line intercepted between the axes is \((1,1)\) then the equation of the line is
1 \(2 x+y=3\)
2 \(2 x-y=1\)
3 \(x-y=0\)
4 \(x+y=2\)
Explanation:
(D) : It is given that \((1,1)\) is the midpoint of line \(\mathrm{AB}\).
\(\Rightarrow 1=\frac{a+0}{2}, 1=\frac{0+\mathrm{b}}{2}\)
\(\Rightarrow a=2 \Rightarrow b=2\)
Equation of line \(\mathrm{AB}\) is
\(\frac{\mathrm{x}}{2}+\frac{\mathrm{y}}{2}=1 \Rightarrow \mathrm{x}+\mathrm{y}=2\)
COMEDK-2011
Straight Line
88648
The perpendicular distance of a line from the origin is \(7 \mathrm{~cm}\) and its slope is -1 . The equation of line is
1 \(x+y+7 \sqrt{2}=0\)
2 \(x-y+7 \sqrt{2}=0\)
3 \(x+y-7 \sqrt{2}=0\)
4 \(x-y-7 \sqrt{2}=0\)
Explanation:
(C) : Equation of line in slope form is\(\mathrm{y}=\mathrm{mx}+\mathrm{c}\)
We have, slope \(=-1\)
So, equation is \(\mathrm{y}=-\mathrm{x}+\mathrm{c}\)
\(\mathrm{x}+\mathrm{y}-\mathrm{c}=0\)
Perpendicular distance of this line from
Origin is \(=7 \mathrm{~cm}\).
Then,
Perpendicular distance from the origin \((0,0)\) to the
Line, \(\quad \mathrm{x}+\mathrm{y}-\mathrm{c}=0\) is-
\(7 =\left|\frac{0+0-c}{\sqrt{2}}\right|\)
\(7 \sqrt{2} =c\)
So, equation are-
\(x-y-7 \sqrt{2}=0\)
88672
A square is formed by the lines \(x=0, y=0, x=1\), \(y=1\). Then, the equation of its diagonals will be
1 \(y=x, x+y=2\)
2 \(2 \mathrm{y}=x, x+y=\frac{1}{2}\)
3 \(\mathrm{y}=x, x+y=1\)
4 \(y=2 x, z+y=\frac{1}{4}\)
Explanation:
(C) : Equation of \(\mathrm{OB}\)
\(y=\left(\frac{1-0}{1-0}\right) x\)
\(\Rightarrow \quad(\mathrm{y}=\mathrm{x})\)
Equation of \(\mathrm{AC}\)
\(y-0=-1(x-1)\)
\(\Rightarrow \quad y=-x+1 \Rightarrow x+y=1\)
AP EAMCET-2020-17.09.2020
Straight Line
88673
Find the value of ' \(k\) ' if the angle between the straight lines represented by
\(2 x^{2}+5 x y+3 y^{2}+6 x+7 y+4=0 \text { is } \tan ^{-1}(k) .\)
1 \(\left(\frac{1}{5}\right)\) only
2 \(\left(\frac{-1}{5}\right)\) only
3 \(\pm \frac{1}{5}\)
4 0
Explanation:
(C) : Given,
Straight lines \(2 x^{2}+5 x y+3 y^{2}+6 x+7 y+4=0\)
Angle between straight lines represented by
\(\tan \theta =\frac{3 \sqrt{\mathrm{h}^{2}-\mathrm{ab}}}{\mathrm{a}+\mathrm{b}}\)
\(\tan \theta =\frac{2 \sqrt{(5 / 2)^{2}-6}}{3}= \pm \frac{1}{5}\)
\(a x^{2}+b y^{2}+2 g x+2 f y+c+2 h x y=0 \text { is }\)
Hence,
\(\theta=\tan ^{-1} \mathrm{k}=\tan ^{-1}\left( \pm \frac{1}{5}\right) \Rightarrow\left(\mathrm{k}= \pm \frac{1}{5}\right)\)
AP EAMCET-2020-17.09.2020
Straight Line
88689
A variable line passes through the fixed point \((\alpha, \beta)\) The locus of the foot of the perpendicular from the origin on the line is
1 \(x^{2}+y^{2}-\alpha x-\beta y=0\)
2 \(x^{2}+y^{2}+2 \alpha x+2 \beta y=0\)
3 \(\alpha x+\beta y \pm \sqrt{\left(\alpha^{2}+\beta^{2}\right)}=0\)
(A) : Let \((\alpha, \beta)\) be the given point, let \(Q(x, y)\) be the foot of the perpendicular, and let \(\mathrm{O}\) be the origin.
The line can have any direction
\(\angle \mathrm{PQO}=90^{\circ}\)
Point \(\mathrm{Q}\) lies on the circle having diameter \(\mathrm{OP}\), The locus of point \(\mathrm{Q}\).
\((x-0)(x-\alpha)+(y-0)(y-\beta)=0\)
\(\Rightarrow \quad x^{2}-x \alpha+y^{2}-y \beta=0\)
\(\Rightarrow \quad x^{2}+y^{2}-\alpha x-\beta y=0\)
WB JEE-2019
Straight Line
88647
If the midpoint of the section of the straight line intercepted between the axes is \((1,1)\) then the equation of the line is
1 \(2 x+y=3\)
2 \(2 x-y=1\)
3 \(x-y=0\)
4 \(x+y=2\)
Explanation:
(D) : It is given that \((1,1)\) is the midpoint of line \(\mathrm{AB}\).
\(\Rightarrow 1=\frac{a+0}{2}, 1=\frac{0+\mathrm{b}}{2}\)
\(\Rightarrow a=2 \Rightarrow b=2\)
Equation of line \(\mathrm{AB}\) is
\(\frac{\mathrm{x}}{2}+\frac{\mathrm{y}}{2}=1 \Rightarrow \mathrm{x}+\mathrm{y}=2\)
COMEDK-2011
Straight Line
88648
The perpendicular distance of a line from the origin is \(7 \mathrm{~cm}\) and its slope is -1 . The equation of line is
1 \(x+y+7 \sqrt{2}=0\)
2 \(x-y+7 \sqrt{2}=0\)
3 \(x+y-7 \sqrt{2}=0\)
4 \(x-y-7 \sqrt{2}=0\)
Explanation:
(C) : Equation of line in slope form is\(\mathrm{y}=\mathrm{mx}+\mathrm{c}\)
We have, slope \(=-1\)
So, equation is \(\mathrm{y}=-\mathrm{x}+\mathrm{c}\)
\(\mathrm{x}+\mathrm{y}-\mathrm{c}=0\)
Perpendicular distance of this line from
Origin is \(=7 \mathrm{~cm}\).
Then,
Perpendicular distance from the origin \((0,0)\) to the
Line, \(\quad \mathrm{x}+\mathrm{y}-\mathrm{c}=0\) is-
\(7 =\left|\frac{0+0-c}{\sqrt{2}}\right|\)
\(7 \sqrt{2} =c\)
So, equation are-
\(x-y-7 \sqrt{2}=0\)
88672
A square is formed by the lines \(x=0, y=0, x=1\), \(y=1\). Then, the equation of its diagonals will be
1 \(y=x, x+y=2\)
2 \(2 \mathrm{y}=x, x+y=\frac{1}{2}\)
3 \(\mathrm{y}=x, x+y=1\)
4 \(y=2 x, z+y=\frac{1}{4}\)
Explanation:
(C) : Equation of \(\mathrm{OB}\)
\(y=\left(\frac{1-0}{1-0}\right) x\)
\(\Rightarrow \quad(\mathrm{y}=\mathrm{x})\)
Equation of \(\mathrm{AC}\)
\(y-0=-1(x-1)\)
\(\Rightarrow \quad y=-x+1 \Rightarrow x+y=1\)
AP EAMCET-2020-17.09.2020
Straight Line
88673
Find the value of ' \(k\) ' if the angle between the straight lines represented by
\(2 x^{2}+5 x y+3 y^{2}+6 x+7 y+4=0 \text { is } \tan ^{-1}(k) .\)
1 \(\left(\frac{1}{5}\right)\) only
2 \(\left(\frac{-1}{5}\right)\) only
3 \(\pm \frac{1}{5}\)
4 0
Explanation:
(C) : Given,
Straight lines \(2 x^{2}+5 x y+3 y^{2}+6 x+7 y+4=0\)
Angle between straight lines represented by
\(\tan \theta =\frac{3 \sqrt{\mathrm{h}^{2}-\mathrm{ab}}}{\mathrm{a}+\mathrm{b}}\)
\(\tan \theta =\frac{2 \sqrt{(5 / 2)^{2}-6}}{3}= \pm \frac{1}{5}\)
\(a x^{2}+b y^{2}+2 g x+2 f y+c+2 h x y=0 \text { is }\)
Hence,
\(\theta=\tan ^{-1} \mathrm{k}=\tan ^{-1}\left( \pm \frac{1}{5}\right) \Rightarrow\left(\mathrm{k}= \pm \frac{1}{5}\right)\)
AP EAMCET-2020-17.09.2020
Straight Line
88689
A variable line passes through the fixed point \((\alpha, \beta)\) The locus of the foot of the perpendicular from the origin on the line is
1 \(x^{2}+y^{2}-\alpha x-\beta y=0\)
2 \(x^{2}+y^{2}+2 \alpha x+2 \beta y=0\)
3 \(\alpha x+\beta y \pm \sqrt{\left(\alpha^{2}+\beta^{2}\right)}=0\)
(A) : Let \((\alpha, \beta)\) be the given point, let \(Q(x, y)\) be the foot of the perpendicular, and let \(\mathrm{O}\) be the origin.
The line can have any direction
\(\angle \mathrm{PQO}=90^{\circ}\)
Point \(\mathrm{Q}\) lies on the circle having diameter \(\mathrm{OP}\), The locus of point \(\mathrm{Q}\).
\((x-0)(x-\alpha)+(y-0)(y-\beta)=0\)
\(\Rightarrow \quad x^{2}-x \alpha+y^{2}-y \beta=0\)
\(\Rightarrow \quad x^{2}+y^{2}-\alpha x-\beta y=0\)
WB JEE-2019
Straight Line
88647
If the midpoint of the section of the straight line intercepted between the axes is \((1,1)\) then the equation of the line is
1 \(2 x+y=3\)
2 \(2 x-y=1\)
3 \(x-y=0\)
4 \(x+y=2\)
Explanation:
(D) : It is given that \((1,1)\) is the midpoint of line \(\mathrm{AB}\).
\(\Rightarrow 1=\frac{a+0}{2}, 1=\frac{0+\mathrm{b}}{2}\)
\(\Rightarrow a=2 \Rightarrow b=2\)
Equation of line \(\mathrm{AB}\) is
\(\frac{\mathrm{x}}{2}+\frac{\mathrm{y}}{2}=1 \Rightarrow \mathrm{x}+\mathrm{y}=2\)
COMEDK-2011
Straight Line
88648
The perpendicular distance of a line from the origin is \(7 \mathrm{~cm}\) and its slope is -1 . The equation of line is
1 \(x+y+7 \sqrt{2}=0\)
2 \(x-y+7 \sqrt{2}=0\)
3 \(x+y-7 \sqrt{2}=0\)
4 \(x-y-7 \sqrt{2}=0\)
Explanation:
(C) : Equation of line in slope form is\(\mathrm{y}=\mathrm{mx}+\mathrm{c}\)
We have, slope \(=-1\)
So, equation is \(\mathrm{y}=-\mathrm{x}+\mathrm{c}\)
\(\mathrm{x}+\mathrm{y}-\mathrm{c}=0\)
Perpendicular distance of this line from
Origin is \(=7 \mathrm{~cm}\).
Then,
Perpendicular distance from the origin \((0,0)\) to the
Line, \(\quad \mathrm{x}+\mathrm{y}-\mathrm{c}=0\) is-
\(7 =\left|\frac{0+0-c}{\sqrt{2}}\right|\)
\(7 \sqrt{2} =c\)
So, equation are-
\(x-y-7 \sqrt{2}=0\)
