88629
The number of values of \(c\) such that the straight line \(y=4 x+c\) touches the curve \(\frac{x^{2}}{4}+y^{2}=1\) is
1 0
2 1
3 2
4 infinite
Explanation:
(C) : Given the straight line \(y=4 x+c\) touches the ellipse \(\frac{x^{2}}{4}+y^{2}=1\)
Substitute \(y=4 x+c\) into ellipse
\(\therefore \quad \frac{\mathrm{x}^{2}}{4}+(4 \mathrm{x}+\mathrm{c})^{2}=1\)
\(\frac{x^{2}}{4}+16 x^{2}+c^{2}+8 c x=1\)
\(x^{2}+64 x^{2}+4 c^{2}+32 c x=4\)
\(65 x^{2}+32 c x+4 c^{2}-4=0\)
The discriminate of this quadratic should be zero
\(\because\) we have only one value of \(x\)
\((32 c)^{2}=4 \times 65 \times\left(4 c^{2}-4\right)\)
\(1024 c^{2}=16 \times 65 \times\left(c^{2}-1\right)\)
\(64 c^{2}=65 \times\left(c^{2}-1\right)\)
\(64 c^{2}=65 c^{2}-65\)
\(c^{2}=65\)
\(c= \pm 65\)
Thus we have two value of \(\mathrm{c}\).
SRM JEE-2014
Straight Line
88630
If the line \(\frac{x}{a}+\frac{y}{b}=1\) moves in such a way that \(\frac{1}{\mathrm{a}^{2}}+\frac{1}{\mathrm{~b}^{2}}=\frac{1}{\mathrm{c}^{2}}\) where, \(\mathrm{c}\) is a constant, then the locus of the foot of perpendicular from the origin to the straight line is-
1 Straight line
2 Parabola
3 Ellipse
4 Circle
Explanation:
(D) :
The equation of line is \(\frac{x}{a}+\frac{y}{b}=1\)
Where,
\(\frac{1}{a^{2}}+\frac{1}{b^{2}}=\frac{1}{c^{2}} \tag{ii}\)
Any line perpendicular to (i) is -
\(\frac{\mathrm{x}}{\mathrm{b}}-\frac{\mathrm{y}}{\mathrm{a}}+\mathrm{k}=0\)
If it is passes through the origin then \(\mathrm{k}=0\)
\(\therefore \quad\) Equation of the line through the origin and perpendicular to (i) is -
\(\frac{\mathrm{x}}{\mathrm{b}}-\frac{\mathrm{y}}{\mathrm{a}}=0 \tag{iii}\)
The locus of the foot of perpendicular from the origin on (i) i.e. locus of the point on intersection of (i) and (iii) is obtained by eliminating the parameter \(\mathrm{a}\) and \(\mathrm{b}\) between them. squaring (1) and (3) and adding, we get-
\(\left(\frac{1}{a^{2}}+\frac{1}{b^{2}}\right) x^{2}+\left(\frac{1}{b^{2}}+\frac{1}{a^{2}}\right) y^{2}=1\)
\(\Rightarrow \frac{\mathrm{x}^{2}}{\mathrm{c}^{2}}+\frac{\mathrm{y}^{2}}{\mathrm{c}^{2}}=1\)
[Using(ii)]
\(\Rightarrow \mathrm{x}^{2}+\mathrm{y}^{2}=\mathrm{c}^{2}\), which is clearly a circle with center at the origin and radius \(\mathrm{c}\).
Manipal-2013
Straight Line
88631
Equation of the straight line passing through \((4,3)\) and making intercepts on the co-ordinate axes whose sum is -1 is
1 \(\frac{x}{2}+\frac{y}{-3}=1\) and \(\frac{x}{-2}+\frac{y}{1}=1\)
2 \(\frac{x}{2}-\frac{y}{3}=-1\) and \(\frac{x}{-2}+\frac{y}{1}=-1\)
3 \(\frac{x}{2}+\frac{y}{3}=1\) and \(\frac{x}{-2}+\frac{y}{-1}=1\)
4 \(\frac{x}{-2}+\frac{y}{3}=-1\) and \(\frac{x}{2}-\frac{y}{1}=1\)
Explanation:
(A) : Let equation of the line are \(\frac{x}{a}+\frac{y}{b}=1\)
Given that, \(\quad \mathrm{a}+\mathrm{b}=-1 \Rightarrow \mathrm{a}=-1-\mathrm{b}\)
Now, putting value of a in (i), we get
\(\frac{x}{-1-b}+\frac{y}{b}=1 \Rightarrow b x+(-1-b) y=b(-1-b)\)
Since the line (ii) passes through \((4,3)\).
\(\therefore \quad 4 b-3-3 b=-b-b^{2}\)
\(b^{2}+2 b-3=0\)
\(b^{2}+3 b-b-3=0\)
\(b(b+3)-1(b+3)=0\)
\((b+3)(b-1)=0\)
\(b=-3 \quad \text { or } \quad b=1\)
Now, if \(b=-3\) then \(a=2\) and if \(b=1\) then \(a=-2\)
\(\therefore\) Equation of line is
\(\frac{\mathrm{x}}{-2}+\frac{\mathrm{y}}{1}=1 \quad\) and \(\quad \frac{\mathrm{x}}{2}+\frac{\mathrm{y}}{-3}=1\)
SRM JEE-2016
Straight Line
88632
An equation of a straight line passing through the intersection of the straight lines \(3 x-4 y+1\) \(=0\) and \(5 x+y-1=0\) and making non-zero, equal intercepts on the axes is
1 \(22 x+22 y=13\)
2 \(23 x+23 y=11\)
3 \(11 x+11 y=23\)
4 \(8 x-3 y=0\)
Explanation:
(B) : Given, \(3 x-4 y+1=0\)
on Solving equation (i) \& (ii), we get
\(x=\frac{3}{23}, y=\frac{8}{23} \tag{ii}\)
Let equation of the required line be \(\frac{x}{a}+\frac{y}{b}=1\)
It passes through \(\left(\frac{3}{23}, \frac{8}{23}\right)\)
\(\therefore \frac{3}{23 a}+\frac{8}{23 a}=1 \Rightarrow 23 a=11 \Rightarrow a=\frac{11}{23}\) \(\therefore\) So, equation of line is \(\frac{23 x}{11}+\frac{23 y}{11}=1\)
\(\Rightarrow 23 \mathrm{x}+23 \mathrm{y}=11\)
88629
The number of values of \(c\) such that the straight line \(y=4 x+c\) touches the curve \(\frac{x^{2}}{4}+y^{2}=1\) is
1 0
2 1
3 2
4 infinite
Explanation:
(C) : Given the straight line \(y=4 x+c\) touches the ellipse \(\frac{x^{2}}{4}+y^{2}=1\)
Substitute \(y=4 x+c\) into ellipse
\(\therefore \quad \frac{\mathrm{x}^{2}}{4}+(4 \mathrm{x}+\mathrm{c})^{2}=1\)
\(\frac{x^{2}}{4}+16 x^{2}+c^{2}+8 c x=1\)
\(x^{2}+64 x^{2}+4 c^{2}+32 c x=4\)
\(65 x^{2}+32 c x+4 c^{2}-4=0\)
The discriminate of this quadratic should be zero
\(\because\) we have only one value of \(x\)
\((32 c)^{2}=4 \times 65 \times\left(4 c^{2}-4\right)\)
\(1024 c^{2}=16 \times 65 \times\left(c^{2}-1\right)\)
\(64 c^{2}=65 \times\left(c^{2}-1\right)\)
\(64 c^{2}=65 c^{2}-65\)
\(c^{2}=65\)
\(c= \pm 65\)
Thus we have two value of \(\mathrm{c}\).
SRM JEE-2014
Straight Line
88630
If the line \(\frac{x}{a}+\frac{y}{b}=1\) moves in such a way that \(\frac{1}{\mathrm{a}^{2}}+\frac{1}{\mathrm{~b}^{2}}=\frac{1}{\mathrm{c}^{2}}\) where, \(\mathrm{c}\) is a constant, then the locus of the foot of perpendicular from the origin to the straight line is-
1 Straight line
2 Parabola
3 Ellipse
4 Circle
Explanation:
(D) :
The equation of line is \(\frac{x}{a}+\frac{y}{b}=1\)
Where,
\(\frac{1}{a^{2}}+\frac{1}{b^{2}}=\frac{1}{c^{2}} \tag{ii}\)
Any line perpendicular to (i) is -
\(\frac{\mathrm{x}}{\mathrm{b}}-\frac{\mathrm{y}}{\mathrm{a}}+\mathrm{k}=0\)
If it is passes through the origin then \(\mathrm{k}=0\)
\(\therefore \quad\) Equation of the line through the origin and perpendicular to (i) is -
\(\frac{\mathrm{x}}{\mathrm{b}}-\frac{\mathrm{y}}{\mathrm{a}}=0 \tag{iii}\)
The locus of the foot of perpendicular from the origin on (i) i.e. locus of the point on intersection of (i) and (iii) is obtained by eliminating the parameter \(\mathrm{a}\) and \(\mathrm{b}\) between them. squaring (1) and (3) and adding, we get-
\(\left(\frac{1}{a^{2}}+\frac{1}{b^{2}}\right) x^{2}+\left(\frac{1}{b^{2}}+\frac{1}{a^{2}}\right) y^{2}=1\)
\(\Rightarrow \frac{\mathrm{x}^{2}}{\mathrm{c}^{2}}+\frac{\mathrm{y}^{2}}{\mathrm{c}^{2}}=1\)
[Using(ii)]
\(\Rightarrow \mathrm{x}^{2}+\mathrm{y}^{2}=\mathrm{c}^{2}\), which is clearly a circle with center at the origin and radius \(\mathrm{c}\).
Manipal-2013
Straight Line
88631
Equation of the straight line passing through \((4,3)\) and making intercepts on the co-ordinate axes whose sum is -1 is
1 \(\frac{x}{2}+\frac{y}{-3}=1\) and \(\frac{x}{-2}+\frac{y}{1}=1\)
2 \(\frac{x}{2}-\frac{y}{3}=-1\) and \(\frac{x}{-2}+\frac{y}{1}=-1\)
3 \(\frac{x}{2}+\frac{y}{3}=1\) and \(\frac{x}{-2}+\frac{y}{-1}=1\)
4 \(\frac{x}{-2}+\frac{y}{3}=-1\) and \(\frac{x}{2}-\frac{y}{1}=1\)
Explanation:
(A) : Let equation of the line are \(\frac{x}{a}+\frac{y}{b}=1\)
Given that, \(\quad \mathrm{a}+\mathrm{b}=-1 \Rightarrow \mathrm{a}=-1-\mathrm{b}\)
Now, putting value of a in (i), we get
\(\frac{x}{-1-b}+\frac{y}{b}=1 \Rightarrow b x+(-1-b) y=b(-1-b)\)
Since the line (ii) passes through \((4,3)\).
\(\therefore \quad 4 b-3-3 b=-b-b^{2}\)
\(b^{2}+2 b-3=0\)
\(b^{2}+3 b-b-3=0\)
\(b(b+3)-1(b+3)=0\)
\((b+3)(b-1)=0\)
\(b=-3 \quad \text { or } \quad b=1\)
Now, if \(b=-3\) then \(a=2\) and if \(b=1\) then \(a=-2\)
\(\therefore\) Equation of line is
\(\frac{\mathrm{x}}{-2}+\frac{\mathrm{y}}{1}=1 \quad\) and \(\quad \frac{\mathrm{x}}{2}+\frac{\mathrm{y}}{-3}=1\)
SRM JEE-2016
Straight Line
88632
An equation of a straight line passing through the intersection of the straight lines \(3 x-4 y+1\) \(=0\) and \(5 x+y-1=0\) and making non-zero, equal intercepts on the axes is
1 \(22 x+22 y=13\)
2 \(23 x+23 y=11\)
3 \(11 x+11 y=23\)
4 \(8 x-3 y=0\)
Explanation:
(B) : Given, \(3 x-4 y+1=0\)
on Solving equation (i) \& (ii), we get
\(x=\frac{3}{23}, y=\frac{8}{23} \tag{ii}\)
Let equation of the required line be \(\frac{x}{a}+\frac{y}{b}=1\)
It passes through \(\left(\frac{3}{23}, \frac{8}{23}\right)\)
\(\therefore \frac{3}{23 a}+\frac{8}{23 a}=1 \Rightarrow 23 a=11 \Rightarrow a=\frac{11}{23}\) \(\therefore\) So, equation of line is \(\frac{23 x}{11}+\frac{23 y}{11}=1\)
\(\Rightarrow 23 \mathrm{x}+23 \mathrm{y}=11\)
88629
The number of values of \(c\) such that the straight line \(y=4 x+c\) touches the curve \(\frac{x^{2}}{4}+y^{2}=1\) is
1 0
2 1
3 2
4 infinite
Explanation:
(C) : Given the straight line \(y=4 x+c\) touches the ellipse \(\frac{x^{2}}{4}+y^{2}=1\)
Substitute \(y=4 x+c\) into ellipse
\(\therefore \quad \frac{\mathrm{x}^{2}}{4}+(4 \mathrm{x}+\mathrm{c})^{2}=1\)
\(\frac{x^{2}}{4}+16 x^{2}+c^{2}+8 c x=1\)
\(x^{2}+64 x^{2}+4 c^{2}+32 c x=4\)
\(65 x^{2}+32 c x+4 c^{2}-4=0\)
The discriminate of this quadratic should be zero
\(\because\) we have only one value of \(x\)
\((32 c)^{2}=4 \times 65 \times\left(4 c^{2}-4\right)\)
\(1024 c^{2}=16 \times 65 \times\left(c^{2}-1\right)\)
\(64 c^{2}=65 \times\left(c^{2}-1\right)\)
\(64 c^{2}=65 c^{2}-65\)
\(c^{2}=65\)
\(c= \pm 65\)
Thus we have two value of \(\mathrm{c}\).
SRM JEE-2014
Straight Line
88630
If the line \(\frac{x}{a}+\frac{y}{b}=1\) moves in such a way that \(\frac{1}{\mathrm{a}^{2}}+\frac{1}{\mathrm{~b}^{2}}=\frac{1}{\mathrm{c}^{2}}\) where, \(\mathrm{c}\) is a constant, then the locus of the foot of perpendicular from the origin to the straight line is-
1 Straight line
2 Parabola
3 Ellipse
4 Circle
Explanation:
(D) :
The equation of line is \(\frac{x}{a}+\frac{y}{b}=1\)
Where,
\(\frac{1}{a^{2}}+\frac{1}{b^{2}}=\frac{1}{c^{2}} \tag{ii}\)
Any line perpendicular to (i) is -
\(\frac{\mathrm{x}}{\mathrm{b}}-\frac{\mathrm{y}}{\mathrm{a}}+\mathrm{k}=0\)
If it is passes through the origin then \(\mathrm{k}=0\)
\(\therefore \quad\) Equation of the line through the origin and perpendicular to (i) is -
\(\frac{\mathrm{x}}{\mathrm{b}}-\frac{\mathrm{y}}{\mathrm{a}}=0 \tag{iii}\)
The locus of the foot of perpendicular from the origin on (i) i.e. locus of the point on intersection of (i) and (iii) is obtained by eliminating the parameter \(\mathrm{a}\) and \(\mathrm{b}\) between them. squaring (1) and (3) and adding, we get-
\(\left(\frac{1}{a^{2}}+\frac{1}{b^{2}}\right) x^{2}+\left(\frac{1}{b^{2}}+\frac{1}{a^{2}}\right) y^{2}=1\)
\(\Rightarrow \frac{\mathrm{x}^{2}}{\mathrm{c}^{2}}+\frac{\mathrm{y}^{2}}{\mathrm{c}^{2}}=1\)
[Using(ii)]
\(\Rightarrow \mathrm{x}^{2}+\mathrm{y}^{2}=\mathrm{c}^{2}\), which is clearly a circle with center at the origin and radius \(\mathrm{c}\).
Manipal-2013
Straight Line
88631
Equation of the straight line passing through \((4,3)\) and making intercepts on the co-ordinate axes whose sum is -1 is
1 \(\frac{x}{2}+\frac{y}{-3}=1\) and \(\frac{x}{-2}+\frac{y}{1}=1\)
2 \(\frac{x}{2}-\frac{y}{3}=-1\) and \(\frac{x}{-2}+\frac{y}{1}=-1\)
3 \(\frac{x}{2}+\frac{y}{3}=1\) and \(\frac{x}{-2}+\frac{y}{-1}=1\)
4 \(\frac{x}{-2}+\frac{y}{3}=-1\) and \(\frac{x}{2}-\frac{y}{1}=1\)
Explanation:
(A) : Let equation of the line are \(\frac{x}{a}+\frac{y}{b}=1\)
Given that, \(\quad \mathrm{a}+\mathrm{b}=-1 \Rightarrow \mathrm{a}=-1-\mathrm{b}\)
Now, putting value of a in (i), we get
\(\frac{x}{-1-b}+\frac{y}{b}=1 \Rightarrow b x+(-1-b) y=b(-1-b)\)
Since the line (ii) passes through \((4,3)\).
\(\therefore \quad 4 b-3-3 b=-b-b^{2}\)
\(b^{2}+2 b-3=0\)
\(b^{2}+3 b-b-3=0\)
\(b(b+3)-1(b+3)=0\)
\((b+3)(b-1)=0\)
\(b=-3 \quad \text { or } \quad b=1\)
Now, if \(b=-3\) then \(a=2\) and if \(b=1\) then \(a=-2\)
\(\therefore\) Equation of line is
\(\frac{\mathrm{x}}{-2}+\frac{\mathrm{y}}{1}=1 \quad\) and \(\quad \frac{\mathrm{x}}{2}+\frac{\mathrm{y}}{-3}=1\)
SRM JEE-2016
Straight Line
88632
An equation of a straight line passing through the intersection of the straight lines \(3 x-4 y+1\) \(=0\) and \(5 x+y-1=0\) and making non-zero, equal intercepts on the axes is
1 \(22 x+22 y=13\)
2 \(23 x+23 y=11\)
3 \(11 x+11 y=23\)
4 \(8 x-3 y=0\)
Explanation:
(B) : Given, \(3 x-4 y+1=0\)
on Solving equation (i) \& (ii), we get
\(x=\frac{3}{23}, y=\frac{8}{23} \tag{ii}\)
Let equation of the required line be \(\frac{x}{a}+\frac{y}{b}=1\)
It passes through \(\left(\frac{3}{23}, \frac{8}{23}\right)\)
\(\therefore \frac{3}{23 a}+\frac{8}{23 a}=1 \Rightarrow 23 a=11 \Rightarrow a=\frac{11}{23}\) \(\therefore\) So, equation of line is \(\frac{23 x}{11}+\frac{23 y}{11}=1\)
\(\Rightarrow 23 \mathrm{x}+23 \mathrm{y}=11\)
88629
The number of values of \(c\) such that the straight line \(y=4 x+c\) touches the curve \(\frac{x^{2}}{4}+y^{2}=1\) is
1 0
2 1
3 2
4 infinite
Explanation:
(C) : Given the straight line \(y=4 x+c\) touches the ellipse \(\frac{x^{2}}{4}+y^{2}=1\)
Substitute \(y=4 x+c\) into ellipse
\(\therefore \quad \frac{\mathrm{x}^{2}}{4}+(4 \mathrm{x}+\mathrm{c})^{2}=1\)
\(\frac{x^{2}}{4}+16 x^{2}+c^{2}+8 c x=1\)
\(x^{2}+64 x^{2}+4 c^{2}+32 c x=4\)
\(65 x^{2}+32 c x+4 c^{2}-4=0\)
The discriminate of this quadratic should be zero
\(\because\) we have only one value of \(x\)
\((32 c)^{2}=4 \times 65 \times\left(4 c^{2}-4\right)\)
\(1024 c^{2}=16 \times 65 \times\left(c^{2}-1\right)\)
\(64 c^{2}=65 \times\left(c^{2}-1\right)\)
\(64 c^{2}=65 c^{2}-65\)
\(c^{2}=65\)
\(c= \pm 65\)
Thus we have two value of \(\mathrm{c}\).
SRM JEE-2014
Straight Line
88630
If the line \(\frac{x}{a}+\frac{y}{b}=1\) moves in such a way that \(\frac{1}{\mathrm{a}^{2}}+\frac{1}{\mathrm{~b}^{2}}=\frac{1}{\mathrm{c}^{2}}\) where, \(\mathrm{c}\) is a constant, then the locus of the foot of perpendicular from the origin to the straight line is-
1 Straight line
2 Parabola
3 Ellipse
4 Circle
Explanation:
(D) :
The equation of line is \(\frac{x}{a}+\frac{y}{b}=1\)
Where,
\(\frac{1}{a^{2}}+\frac{1}{b^{2}}=\frac{1}{c^{2}} \tag{ii}\)
Any line perpendicular to (i) is -
\(\frac{\mathrm{x}}{\mathrm{b}}-\frac{\mathrm{y}}{\mathrm{a}}+\mathrm{k}=0\)
If it is passes through the origin then \(\mathrm{k}=0\)
\(\therefore \quad\) Equation of the line through the origin and perpendicular to (i) is -
\(\frac{\mathrm{x}}{\mathrm{b}}-\frac{\mathrm{y}}{\mathrm{a}}=0 \tag{iii}\)
The locus of the foot of perpendicular from the origin on (i) i.e. locus of the point on intersection of (i) and (iii) is obtained by eliminating the parameter \(\mathrm{a}\) and \(\mathrm{b}\) between them. squaring (1) and (3) and adding, we get-
\(\left(\frac{1}{a^{2}}+\frac{1}{b^{2}}\right) x^{2}+\left(\frac{1}{b^{2}}+\frac{1}{a^{2}}\right) y^{2}=1\)
\(\Rightarrow \frac{\mathrm{x}^{2}}{\mathrm{c}^{2}}+\frac{\mathrm{y}^{2}}{\mathrm{c}^{2}}=1\)
[Using(ii)]
\(\Rightarrow \mathrm{x}^{2}+\mathrm{y}^{2}=\mathrm{c}^{2}\), which is clearly a circle with center at the origin and radius \(\mathrm{c}\).
Manipal-2013
Straight Line
88631
Equation of the straight line passing through \((4,3)\) and making intercepts on the co-ordinate axes whose sum is -1 is
1 \(\frac{x}{2}+\frac{y}{-3}=1\) and \(\frac{x}{-2}+\frac{y}{1}=1\)
2 \(\frac{x}{2}-\frac{y}{3}=-1\) and \(\frac{x}{-2}+\frac{y}{1}=-1\)
3 \(\frac{x}{2}+\frac{y}{3}=1\) and \(\frac{x}{-2}+\frac{y}{-1}=1\)
4 \(\frac{x}{-2}+\frac{y}{3}=-1\) and \(\frac{x}{2}-\frac{y}{1}=1\)
Explanation:
(A) : Let equation of the line are \(\frac{x}{a}+\frac{y}{b}=1\)
Given that, \(\quad \mathrm{a}+\mathrm{b}=-1 \Rightarrow \mathrm{a}=-1-\mathrm{b}\)
Now, putting value of a in (i), we get
\(\frac{x}{-1-b}+\frac{y}{b}=1 \Rightarrow b x+(-1-b) y=b(-1-b)\)
Since the line (ii) passes through \((4,3)\).
\(\therefore \quad 4 b-3-3 b=-b-b^{2}\)
\(b^{2}+2 b-3=0\)
\(b^{2}+3 b-b-3=0\)
\(b(b+3)-1(b+3)=0\)
\((b+3)(b-1)=0\)
\(b=-3 \quad \text { or } \quad b=1\)
Now, if \(b=-3\) then \(a=2\) and if \(b=1\) then \(a=-2\)
\(\therefore\) Equation of line is
\(\frac{\mathrm{x}}{-2}+\frac{\mathrm{y}}{1}=1 \quad\) and \(\quad \frac{\mathrm{x}}{2}+\frac{\mathrm{y}}{-3}=1\)
SRM JEE-2016
Straight Line
88632
An equation of a straight line passing through the intersection of the straight lines \(3 x-4 y+1\) \(=0\) and \(5 x+y-1=0\) and making non-zero, equal intercepts on the axes is
1 \(22 x+22 y=13\)
2 \(23 x+23 y=11\)
3 \(11 x+11 y=23\)
4 \(8 x-3 y=0\)
Explanation:
(B) : Given, \(3 x-4 y+1=0\)
on Solving equation (i) \& (ii), we get
\(x=\frac{3}{23}, y=\frac{8}{23} \tag{ii}\)
Let equation of the required line be \(\frac{x}{a}+\frac{y}{b}=1\)
It passes through \(\left(\frac{3}{23}, \frac{8}{23}\right)\)
\(\therefore \frac{3}{23 a}+\frac{8}{23 a}=1 \Rightarrow 23 a=11 \Rightarrow a=\frac{11}{23}\) \(\therefore\) So, equation of line is \(\frac{23 x}{11}+\frac{23 y}{11}=1\)
\(\Rightarrow 23 \mathrm{x}+23 \mathrm{y}=11\)
