88751
If the algebraic sum of the perpendicular distance from the point \((2,0),(0,2)\) and \((1,1)\) to a variable straight line be zero, then the line passes through the point
1 \((-1,1)\)
2 \((1,1)\)
3 \((1,-1)\)
4 \((-1,-1)\)
Explanation:
(B) : Given, the algebraic sum of the perpendicular distance from the points \((2,0),(0,2)\) and \((1,1)\) to this line is zero.
\(\frac{2 a+0+c}{\sqrt{a^{2}+b^{2}}}+\frac{0+2 b+c}{\sqrt{a^{2}+b^{2}}}+\frac{a+b+c}{\sqrt{a^{2}+b^{2}}}=0\)
\(2 a+c+2 b+c+a+b+c=0\)
\(3 a+3 b+3 c=0\)
\(a+b+c=0\)
Then the line passing through the point \((1,1)\).
[JCECE-2011]
Straight Line
88752
Identify the point on the line \(2 x+3 y+7=0\), which is at a distance of +3 units from \((1,3)\)
(C): Let \(\mathrm{P}(\alpha, \beta)\) be the point on the line \(2 \mathrm{x}+3 \mathrm{y}+7=0\)
\(2 \alpha+3 \beta+7=0\)
\(\beta=\left(\frac{-7-2 \alpha}{3}\right)\)
\(\therefore \quad \mathrm{P}=\left(\alpha, \frac{-7-2 \alpha}{3}\right)\)
Given, point \(A=(1,-3)\)
\(\because \quad \mathrm{AP}=3\)
\(\sqrt{(\alpha-1)^{2}+\left(\frac{-7-2 \alpha}{3}+3\right)^{2}}=3\)
\(\alpha^{2}+1-2 \alpha+\frac{49+4 \alpha^{2}+28 \alpha}{9}+9-14-4 \alpha=9\)
\(9 \alpha^{2}+9-18 \alpha+49+4 \alpha^{2}+28 \alpha-126-36 \alpha=0\)
\(13 \alpha^{2}-26 \alpha-68=0\)
\(\alpha=\frac{26 \pm \sqrt{676+3536}}{26}\)
\(\alpha=1 \pm \frac{9}{\sqrt{13}}=\frac{\sqrt{13} \pm 9}{\sqrt{13}}\)
When, \(\alpha=\frac{\sqrt{13}-9}{\sqrt{13}}\)
\(\beta=\frac{-7-2 \alpha}{3}=\frac{-3 \sqrt{13}+6}{\sqrt{13}}\)
So, \(\quad \mathrm{P}=\left(\frac{\sqrt{13}-9}{\sqrt{13}}, \frac{-3 \sqrt{13}+6}{\sqrt{13}}\right)\)
APEAPCET-2021-20.08.2021
Straight Line
88753
The equation of the base of an equilateral triangle is \(x+y=2\) and one vertex is \((2,-1)\), then the length of the side of the triangle is
1 \(\sqrt{3 / 2} / \sqrt{2 / 3}\)
2 \(\sqrt{2}\)
3 \(\sqrt{2 / 3}\)
4 \(\sqrt{3 / 2}\)
Explanation:
(C) :
\(\mathrm{d}=\left|\frac{\mathrm{Ax}+\mathrm{By}+\mathrm{c}}{\sqrt{A^{2}+B^{2}}}\right|=\left|\frac{2-1-2}{\sqrt{1+1}}\right|=\left|\frac{-1}{\sqrt{2}}\right|\)
\(\mathrm{d}=\frac{1}{\sqrt{2}}\)
Suppose the length of the side is \(2 \mathrm{~A}\)
\((2 l)^{2}=(l)^{2}+\left(\frac{1}{\sqrt{2}}\right)^{2}\)
\(4 l^{2}=l^{2}+\frac{1}{2}\)
\(l^{2}=1 / 6\)
\(l=\frac{1}{\sqrt{6}}\)
The length of side of the triangle is
\(2 l=2 \times \frac{1}{\sqrt{6}}=\frac{2}{\sqrt{6}}=\sqrt{\frac{2}{3}}\)
Shift-II]
Straight Line
88754
If a point ' \(P\) ' on the line \(3 x+5 y=15\) is equidistant from the coordinate axes, then \(P\) lies \(3 x+5 y=15\)
1 Only in the first quadrant
2 Either in first or in second quadrant
3 Either in first or in third quadrant
4 Only in the third quadrant
Explanation:
(B):
\(\frac{x}{5}+\frac{y}{3}=1\)
\(|\mathrm{a}|=|\mathrm{b}|\)
\(|\mathrm{a}|=\left|\frac{15-3 \mathrm{a}}{5}\right|\)
\(\mathrm{a}=\frac{15-3 \mathrm{a}}{5} \quad \mathrm{a}=\frac{-15+3 \mathrm{a}}{5}\)
\(\mathrm{a}=\frac{15}{2} 2 \mathrm{a}=-15\)
\(\mathrm{a}=-15 / 2\)
Either in first or in second quadrant.
Shift-II]
Straight Line
88755
What is the perpendicular distance of a corner to the diagonal not passing through that corner?
1 \(\frac{2}{3}\)
2 \(\frac{\sqrt{3}}{2}\)
3 \(\sqrt{\frac{2}{3}}\)
4 None of the above
Explanation:
(C):
Let the edges \(\mathrm{OA}, \mathrm{OB}, \mathrm{OC}\) of unit cube represents the unit vectors \(\hat{i}, \hat{j}, \hat{k}\) resp. Let \(\mathrm{cm}\) be the, perpendicular from corner on diagonal OP.
We have \(\overrightarrow{\mathrm{OP}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}\)
\(|\overrightarrow{\mathrm{OP}}|=|\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}|=\sqrt{3}\)
Unit vector in the direction of \(\overrightarrow{\mathrm{OP}}\) is \(\frac{\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}}{\sqrt{3}}\)
\(\mathrm{OM}=\) projection of \(\mathrm{OC}\) and \(\mathrm{OP}\)
\(\mathrm{OM}=\overrightarrow{\mathrm{OC}}\) (unit vector along \(\overrightarrow{\mathrm{OP}}\) )
\(O M=\hat{\mathrm{j}} \cdot\left(\frac{\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}}{\sqrt{3}}\right)=\frac{1}{\sqrt{3}}\)
Now, \(\mathrm{CM}^{2}=\mathrm{OC}^{2}-\mathrm{OM}^{2}\)
\(=|\overrightarrow{\mathrm{OC}}|^{2}-\mathrm{OM}^{2}\)
\(\mathrm{CM}^{2}=1-\frac{1}{3}=\frac{2}{3}\)
\(\mathrm{CM}=\sqrt{\frac{2}{3}}\)
88751
If the algebraic sum of the perpendicular distance from the point \((2,0),(0,2)\) and \((1,1)\) to a variable straight line be zero, then the line passes through the point
1 \((-1,1)\)
2 \((1,1)\)
3 \((1,-1)\)
4 \((-1,-1)\)
Explanation:
(B) : Given, the algebraic sum of the perpendicular distance from the points \((2,0),(0,2)\) and \((1,1)\) to this line is zero.
\(\frac{2 a+0+c}{\sqrt{a^{2}+b^{2}}}+\frac{0+2 b+c}{\sqrt{a^{2}+b^{2}}}+\frac{a+b+c}{\sqrt{a^{2}+b^{2}}}=0\)
\(2 a+c+2 b+c+a+b+c=0\)
\(3 a+3 b+3 c=0\)
\(a+b+c=0\)
Then the line passing through the point \((1,1)\).
[JCECE-2011]
Straight Line
88752
Identify the point on the line \(2 x+3 y+7=0\), which is at a distance of +3 units from \((1,3)\)
(C): Let \(\mathrm{P}(\alpha, \beta)\) be the point on the line \(2 \mathrm{x}+3 \mathrm{y}+7=0\)
\(2 \alpha+3 \beta+7=0\)
\(\beta=\left(\frac{-7-2 \alpha}{3}\right)\)
\(\therefore \quad \mathrm{P}=\left(\alpha, \frac{-7-2 \alpha}{3}\right)\)
Given, point \(A=(1,-3)\)
\(\because \quad \mathrm{AP}=3\)
\(\sqrt{(\alpha-1)^{2}+\left(\frac{-7-2 \alpha}{3}+3\right)^{2}}=3\)
\(\alpha^{2}+1-2 \alpha+\frac{49+4 \alpha^{2}+28 \alpha}{9}+9-14-4 \alpha=9\)
\(9 \alpha^{2}+9-18 \alpha+49+4 \alpha^{2}+28 \alpha-126-36 \alpha=0\)
\(13 \alpha^{2}-26 \alpha-68=0\)
\(\alpha=\frac{26 \pm \sqrt{676+3536}}{26}\)
\(\alpha=1 \pm \frac{9}{\sqrt{13}}=\frac{\sqrt{13} \pm 9}{\sqrt{13}}\)
When, \(\alpha=\frac{\sqrt{13}-9}{\sqrt{13}}\)
\(\beta=\frac{-7-2 \alpha}{3}=\frac{-3 \sqrt{13}+6}{\sqrt{13}}\)
So, \(\quad \mathrm{P}=\left(\frac{\sqrt{13}-9}{\sqrt{13}}, \frac{-3 \sqrt{13}+6}{\sqrt{13}}\right)\)
APEAPCET-2021-20.08.2021
Straight Line
88753
The equation of the base of an equilateral triangle is \(x+y=2\) and one vertex is \((2,-1)\), then the length of the side of the triangle is
1 \(\sqrt{3 / 2} / \sqrt{2 / 3}\)
2 \(\sqrt{2}\)
3 \(\sqrt{2 / 3}\)
4 \(\sqrt{3 / 2}\)
Explanation:
(C) :
\(\mathrm{d}=\left|\frac{\mathrm{Ax}+\mathrm{By}+\mathrm{c}}{\sqrt{A^{2}+B^{2}}}\right|=\left|\frac{2-1-2}{\sqrt{1+1}}\right|=\left|\frac{-1}{\sqrt{2}}\right|\)
\(\mathrm{d}=\frac{1}{\sqrt{2}}\)
Suppose the length of the side is \(2 \mathrm{~A}\)
\((2 l)^{2}=(l)^{2}+\left(\frac{1}{\sqrt{2}}\right)^{2}\)
\(4 l^{2}=l^{2}+\frac{1}{2}\)
\(l^{2}=1 / 6\)
\(l=\frac{1}{\sqrt{6}}\)
The length of side of the triangle is
\(2 l=2 \times \frac{1}{\sqrt{6}}=\frac{2}{\sqrt{6}}=\sqrt{\frac{2}{3}}\)
Shift-II]
Straight Line
88754
If a point ' \(P\) ' on the line \(3 x+5 y=15\) is equidistant from the coordinate axes, then \(P\) lies \(3 x+5 y=15\)
1 Only in the first quadrant
2 Either in first or in second quadrant
3 Either in first or in third quadrant
4 Only in the third quadrant
Explanation:
(B):
\(\frac{x}{5}+\frac{y}{3}=1\)
\(|\mathrm{a}|=|\mathrm{b}|\)
\(|\mathrm{a}|=\left|\frac{15-3 \mathrm{a}}{5}\right|\)
\(\mathrm{a}=\frac{15-3 \mathrm{a}}{5} \quad \mathrm{a}=\frac{-15+3 \mathrm{a}}{5}\)
\(\mathrm{a}=\frac{15}{2} 2 \mathrm{a}=-15\)
\(\mathrm{a}=-15 / 2\)
Either in first or in second quadrant.
Shift-II]
Straight Line
88755
What is the perpendicular distance of a corner to the diagonal not passing through that corner?
1 \(\frac{2}{3}\)
2 \(\frac{\sqrt{3}}{2}\)
3 \(\sqrt{\frac{2}{3}}\)
4 None of the above
Explanation:
(C):
Let the edges \(\mathrm{OA}, \mathrm{OB}, \mathrm{OC}\) of unit cube represents the unit vectors \(\hat{i}, \hat{j}, \hat{k}\) resp. Let \(\mathrm{cm}\) be the, perpendicular from corner on diagonal OP.
We have \(\overrightarrow{\mathrm{OP}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}\)
\(|\overrightarrow{\mathrm{OP}}|=|\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}|=\sqrt{3}\)
Unit vector in the direction of \(\overrightarrow{\mathrm{OP}}\) is \(\frac{\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}}{\sqrt{3}}\)
\(\mathrm{OM}=\) projection of \(\mathrm{OC}\) and \(\mathrm{OP}\)
\(\mathrm{OM}=\overrightarrow{\mathrm{OC}}\) (unit vector along \(\overrightarrow{\mathrm{OP}}\) )
\(O M=\hat{\mathrm{j}} \cdot\left(\frac{\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}}{\sqrt{3}}\right)=\frac{1}{\sqrt{3}}\)
Now, \(\mathrm{CM}^{2}=\mathrm{OC}^{2}-\mathrm{OM}^{2}\)
\(=|\overrightarrow{\mathrm{OC}}|^{2}-\mathrm{OM}^{2}\)
\(\mathrm{CM}^{2}=1-\frac{1}{3}=\frac{2}{3}\)
\(\mathrm{CM}=\sqrt{\frac{2}{3}}\)
88751
If the algebraic sum of the perpendicular distance from the point \((2,0),(0,2)\) and \((1,1)\) to a variable straight line be zero, then the line passes through the point
1 \((-1,1)\)
2 \((1,1)\)
3 \((1,-1)\)
4 \((-1,-1)\)
Explanation:
(B) : Given, the algebraic sum of the perpendicular distance from the points \((2,0),(0,2)\) and \((1,1)\) to this line is zero.
\(\frac{2 a+0+c}{\sqrt{a^{2}+b^{2}}}+\frac{0+2 b+c}{\sqrt{a^{2}+b^{2}}}+\frac{a+b+c}{\sqrt{a^{2}+b^{2}}}=0\)
\(2 a+c+2 b+c+a+b+c=0\)
\(3 a+3 b+3 c=0\)
\(a+b+c=0\)
Then the line passing through the point \((1,1)\).
[JCECE-2011]
Straight Line
88752
Identify the point on the line \(2 x+3 y+7=0\), which is at a distance of +3 units from \((1,3)\)
(C): Let \(\mathrm{P}(\alpha, \beta)\) be the point on the line \(2 \mathrm{x}+3 \mathrm{y}+7=0\)
\(2 \alpha+3 \beta+7=0\)
\(\beta=\left(\frac{-7-2 \alpha}{3}\right)\)
\(\therefore \quad \mathrm{P}=\left(\alpha, \frac{-7-2 \alpha}{3}\right)\)
Given, point \(A=(1,-3)\)
\(\because \quad \mathrm{AP}=3\)
\(\sqrt{(\alpha-1)^{2}+\left(\frac{-7-2 \alpha}{3}+3\right)^{2}}=3\)
\(\alpha^{2}+1-2 \alpha+\frac{49+4 \alpha^{2}+28 \alpha}{9}+9-14-4 \alpha=9\)
\(9 \alpha^{2}+9-18 \alpha+49+4 \alpha^{2}+28 \alpha-126-36 \alpha=0\)
\(13 \alpha^{2}-26 \alpha-68=0\)
\(\alpha=\frac{26 \pm \sqrt{676+3536}}{26}\)
\(\alpha=1 \pm \frac{9}{\sqrt{13}}=\frac{\sqrt{13} \pm 9}{\sqrt{13}}\)
When, \(\alpha=\frac{\sqrt{13}-9}{\sqrt{13}}\)
\(\beta=\frac{-7-2 \alpha}{3}=\frac{-3 \sqrt{13}+6}{\sqrt{13}}\)
So, \(\quad \mathrm{P}=\left(\frac{\sqrt{13}-9}{\sqrt{13}}, \frac{-3 \sqrt{13}+6}{\sqrt{13}}\right)\)
APEAPCET-2021-20.08.2021
Straight Line
88753
The equation of the base of an equilateral triangle is \(x+y=2\) and one vertex is \((2,-1)\), then the length of the side of the triangle is
1 \(\sqrt{3 / 2} / \sqrt{2 / 3}\)
2 \(\sqrt{2}\)
3 \(\sqrt{2 / 3}\)
4 \(\sqrt{3 / 2}\)
Explanation:
(C) :
\(\mathrm{d}=\left|\frac{\mathrm{Ax}+\mathrm{By}+\mathrm{c}}{\sqrt{A^{2}+B^{2}}}\right|=\left|\frac{2-1-2}{\sqrt{1+1}}\right|=\left|\frac{-1}{\sqrt{2}}\right|\)
\(\mathrm{d}=\frac{1}{\sqrt{2}}\)
Suppose the length of the side is \(2 \mathrm{~A}\)
\((2 l)^{2}=(l)^{2}+\left(\frac{1}{\sqrt{2}}\right)^{2}\)
\(4 l^{2}=l^{2}+\frac{1}{2}\)
\(l^{2}=1 / 6\)
\(l=\frac{1}{\sqrt{6}}\)
The length of side of the triangle is
\(2 l=2 \times \frac{1}{\sqrt{6}}=\frac{2}{\sqrt{6}}=\sqrt{\frac{2}{3}}\)
Shift-II]
Straight Line
88754
If a point ' \(P\) ' on the line \(3 x+5 y=15\) is equidistant from the coordinate axes, then \(P\) lies \(3 x+5 y=15\)
1 Only in the first quadrant
2 Either in first or in second quadrant
3 Either in first or in third quadrant
4 Only in the third quadrant
Explanation:
(B):
\(\frac{x}{5}+\frac{y}{3}=1\)
\(|\mathrm{a}|=|\mathrm{b}|\)
\(|\mathrm{a}|=\left|\frac{15-3 \mathrm{a}}{5}\right|\)
\(\mathrm{a}=\frac{15-3 \mathrm{a}}{5} \quad \mathrm{a}=\frac{-15+3 \mathrm{a}}{5}\)
\(\mathrm{a}=\frac{15}{2} 2 \mathrm{a}=-15\)
\(\mathrm{a}=-15 / 2\)
Either in first or in second quadrant.
Shift-II]
Straight Line
88755
What is the perpendicular distance of a corner to the diagonal not passing through that corner?
1 \(\frac{2}{3}\)
2 \(\frac{\sqrt{3}}{2}\)
3 \(\sqrt{\frac{2}{3}}\)
4 None of the above
Explanation:
(C):
Let the edges \(\mathrm{OA}, \mathrm{OB}, \mathrm{OC}\) of unit cube represents the unit vectors \(\hat{i}, \hat{j}, \hat{k}\) resp. Let \(\mathrm{cm}\) be the, perpendicular from corner on diagonal OP.
We have \(\overrightarrow{\mathrm{OP}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}\)
\(|\overrightarrow{\mathrm{OP}}|=|\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}|=\sqrt{3}\)
Unit vector in the direction of \(\overrightarrow{\mathrm{OP}}\) is \(\frac{\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}}{\sqrt{3}}\)
\(\mathrm{OM}=\) projection of \(\mathrm{OC}\) and \(\mathrm{OP}\)
\(\mathrm{OM}=\overrightarrow{\mathrm{OC}}\) (unit vector along \(\overrightarrow{\mathrm{OP}}\) )
\(O M=\hat{\mathrm{j}} \cdot\left(\frac{\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}}{\sqrt{3}}\right)=\frac{1}{\sqrt{3}}\)
Now, \(\mathrm{CM}^{2}=\mathrm{OC}^{2}-\mathrm{OM}^{2}\)
\(=|\overrightarrow{\mathrm{OC}}|^{2}-\mathrm{OM}^{2}\)
\(\mathrm{CM}^{2}=1-\frac{1}{3}=\frac{2}{3}\)
\(\mathrm{CM}=\sqrt{\frac{2}{3}}\)
88751
If the algebraic sum of the perpendicular distance from the point \((2,0),(0,2)\) and \((1,1)\) to a variable straight line be zero, then the line passes through the point
1 \((-1,1)\)
2 \((1,1)\)
3 \((1,-1)\)
4 \((-1,-1)\)
Explanation:
(B) : Given, the algebraic sum of the perpendicular distance from the points \((2,0),(0,2)\) and \((1,1)\) to this line is zero.
\(\frac{2 a+0+c}{\sqrt{a^{2}+b^{2}}}+\frac{0+2 b+c}{\sqrt{a^{2}+b^{2}}}+\frac{a+b+c}{\sqrt{a^{2}+b^{2}}}=0\)
\(2 a+c+2 b+c+a+b+c=0\)
\(3 a+3 b+3 c=0\)
\(a+b+c=0\)
Then the line passing through the point \((1,1)\).
[JCECE-2011]
Straight Line
88752
Identify the point on the line \(2 x+3 y+7=0\), which is at a distance of +3 units from \((1,3)\)
(C): Let \(\mathrm{P}(\alpha, \beta)\) be the point on the line \(2 \mathrm{x}+3 \mathrm{y}+7=0\)
\(2 \alpha+3 \beta+7=0\)
\(\beta=\left(\frac{-7-2 \alpha}{3}\right)\)
\(\therefore \quad \mathrm{P}=\left(\alpha, \frac{-7-2 \alpha}{3}\right)\)
Given, point \(A=(1,-3)\)
\(\because \quad \mathrm{AP}=3\)
\(\sqrt{(\alpha-1)^{2}+\left(\frac{-7-2 \alpha}{3}+3\right)^{2}}=3\)
\(\alpha^{2}+1-2 \alpha+\frac{49+4 \alpha^{2}+28 \alpha}{9}+9-14-4 \alpha=9\)
\(9 \alpha^{2}+9-18 \alpha+49+4 \alpha^{2}+28 \alpha-126-36 \alpha=0\)
\(13 \alpha^{2}-26 \alpha-68=0\)
\(\alpha=\frac{26 \pm \sqrt{676+3536}}{26}\)
\(\alpha=1 \pm \frac{9}{\sqrt{13}}=\frac{\sqrt{13} \pm 9}{\sqrt{13}}\)
When, \(\alpha=\frac{\sqrt{13}-9}{\sqrt{13}}\)
\(\beta=\frac{-7-2 \alpha}{3}=\frac{-3 \sqrt{13}+6}{\sqrt{13}}\)
So, \(\quad \mathrm{P}=\left(\frac{\sqrt{13}-9}{\sqrt{13}}, \frac{-3 \sqrt{13}+6}{\sqrt{13}}\right)\)
APEAPCET-2021-20.08.2021
Straight Line
88753
The equation of the base of an equilateral triangle is \(x+y=2\) and one vertex is \((2,-1)\), then the length of the side of the triangle is
1 \(\sqrt{3 / 2} / \sqrt{2 / 3}\)
2 \(\sqrt{2}\)
3 \(\sqrt{2 / 3}\)
4 \(\sqrt{3 / 2}\)
Explanation:
(C) :
\(\mathrm{d}=\left|\frac{\mathrm{Ax}+\mathrm{By}+\mathrm{c}}{\sqrt{A^{2}+B^{2}}}\right|=\left|\frac{2-1-2}{\sqrt{1+1}}\right|=\left|\frac{-1}{\sqrt{2}}\right|\)
\(\mathrm{d}=\frac{1}{\sqrt{2}}\)
Suppose the length of the side is \(2 \mathrm{~A}\)
\((2 l)^{2}=(l)^{2}+\left(\frac{1}{\sqrt{2}}\right)^{2}\)
\(4 l^{2}=l^{2}+\frac{1}{2}\)
\(l^{2}=1 / 6\)
\(l=\frac{1}{\sqrt{6}}\)
The length of side of the triangle is
\(2 l=2 \times \frac{1}{\sqrt{6}}=\frac{2}{\sqrt{6}}=\sqrt{\frac{2}{3}}\)
Shift-II]
Straight Line
88754
If a point ' \(P\) ' on the line \(3 x+5 y=15\) is equidistant from the coordinate axes, then \(P\) lies \(3 x+5 y=15\)
1 Only in the first quadrant
2 Either in first or in second quadrant
3 Either in first or in third quadrant
4 Only in the third quadrant
Explanation:
(B):
\(\frac{x}{5}+\frac{y}{3}=1\)
\(|\mathrm{a}|=|\mathrm{b}|\)
\(|\mathrm{a}|=\left|\frac{15-3 \mathrm{a}}{5}\right|\)
\(\mathrm{a}=\frac{15-3 \mathrm{a}}{5} \quad \mathrm{a}=\frac{-15+3 \mathrm{a}}{5}\)
\(\mathrm{a}=\frac{15}{2} 2 \mathrm{a}=-15\)
\(\mathrm{a}=-15 / 2\)
Either in first or in second quadrant.
Shift-II]
Straight Line
88755
What is the perpendicular distance of a corner to the diagonal not passing through that corner?
1 \(\frac{2}{3}\)
2 \(\frac{\sqrt{3}}{2}\)
3 \(\sqrt{\frac{2}{3}}\)
4 None of the above
Explanation:
(C):
Let the edges \(\mathrm{OA}, \mathrm{OB}, \mathrm{OC}\) of unit cube represents the unit vectors \(\hat{i}, \hat{j}, \hat{k}\) resp. Let \(\mathrm{cm}\) be the, perpendicular from corner on diagonal OP.
We have \(\overrightarrow{\mathrm{OP}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}\)
\(|\overrightarrow{\mathrm{OP}}|=|\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}|=\sqrt{3}\)
Unit vector in the direction of \(\overrightarrow{\mathrm{OP}}\) is \(\frac{\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}}{\sqrt{3}}\)
\(\mathrm{OM}=\) projection of \(\mathrm{OC}\) and \(\mathrm{OP}\)
\(\mathrm{OM}=\overrightarrow{\mathrm{OC}}\) (unit vector along \(\overrightarrow{\mathrm{OP}}\) )
\(O M=\hat{\mathrm{j}} \cdot\left(\frac{\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}}{\sqrt{3}}\right)=\frac{1}{\sqrt{3}}\)
Now, \(\mathrm{CM}^{2}=\mathrm{OC}^{2}-\mathrm{OM}^{2}\)
\(=|\overrightarrow{\mathrm{OC}}|^{2}-\mathrm{OM}^{2}\)
\(\mathrm{CM}^{2}=1-\frac{1}{3}=\frac{2}{3}\)
\(\mathrm{CM}=\sqrt{\frac{2}{3}}\)
88751
If the algebraic sum of the perpendicular distance from the point \((2,0),(0,2)\) and \((1,1)\) to a variable straight line be zero, then the line passes through the point
1 \((-1,1)\)
2 \((1,1)\)
3 \((1,-1)\)
4 \((-1,-1)\)
Explanation:
(B) : Given, the algebraic sum of the perpendicular distance from the points \((2,0),(0,2)\) and \((1,1)\) to this line is zero.
\(\frac{2 a+0+c}{\sqrt{a^{2}+b^{2}}}+\frac{0+2 b+c}{\sqrt{a^{2}+b^{2}}}+\frac{a+b+c}{\sqrt{a^{2}+b^{2}}}=0\)
\(2 a+c+2 b+c+a+b+c=0\)
\(3 a+3 b+3 c=0\)
\(a+b+c=0\)
Then the line passing through the point \((1,1)\).
[JCECE-2011]
Straight Line
88752
Identify the point on the line \(2 x+3 y+7=0\), which is at a distance of +3 units from \((1,3)\)
(C): Let \(\mathrm{P}(\alpha, \beta)\) be the point on the line \(2 \mathrm{x}+3 \mathrm{y}+7=0\)
\(2 \alpha+3 \beta+7=0\)
\(\beta=\left(\frac{-7-2 \alpha}{3}\right)\)
\(\therefore \quad \mathrm{P}=\left(\alpha, \frac{-7-2 \alpha}{3}\right)\)
Given, point \(A=(1,-3)\)
\(\because \quad \mathrm{AP}=3\)
\(\sqrt{(\alpha-1)^{2}+\left(\frac{-7-2 \alpha}{3}+3\right)^{2}}=3\)
\(\alpha^{2}+1-2 \alpha+\frac{49+4 \alpha^{2}+28 \alpha}{9}+9-14-4 \alpha=9\)
\(9 \alpha^{2}+9-18 \alpha+49+4 \alpha^{2}+28 \alpha-126-36 \alpha=0\)
\(13 \alpha^{2}-26 \alpha-68=0\)
\(\alpha=\frac{26 \pm \sqrt{676+3536}}{26}\)
\(\alpha=1 \pm \frac{9}{\sqrt{13}}=\frac{\sqrt{13} \pm 9}{\sqrt{13}}\)
When, \(\alpha=\frac{\sqrt{13}-9}{\sqrt{13}}\)
\(\beta=\frac{-7-2 \alpha}{3}=\frac{-3 \sqrt{13}+6}{\sqrt{13}}\)
So, \(\quad \mathrm{P}=\left(\frac{\sqrt{13}-9}{\sqrt{13}}, \frac{-3 \sqrt{13}+6}{\sqrt{13}}\right)\)
APEAPCET-2021-20.08.2021
Straight Line
88753
The equation of the base of an equilateral triangle is \(x+y=2\) and one vertex is \((2,-1)\), then the length of the side of the triangle is
1 \(\sqrt{3 / 2} / \sqrt{2 / 3}\)
2 \(\sqrt{2}\)
3 \(\sqrt{2 / 3}\)
4 \(\sqrt{3 / 2}\)
Explanation:
(C) :
\(\mathrm{d}=\left|\frac{\mathrm{Ax}+\mathrm{By}+\mathrm{c}}{\sqrt{A^{2}+B^{2}}}\right|=\left|\frac{2-1-2}{\sqrt{1+1}}\right|=\left|\frac{-1}{\sqrt{2}}\right|\)
\(\mathrm{d}=\frac{1}{\sqrt{2}}\)
Suppose the length of the side is \(2 \mathrm{~A}\)
\((2 l)^{2}=(l)^{2}+\left(\frac{1}{\sqrt{2}}\right)^{2}\)
\(4 l^{2}=l^{2}+\frac{1}{2}\)
\(l^{2}=1 / 6\)
\(l=\frac{1}{\sqrt{6}}\)
The length of side of the triangle is
\(2 l=2 \times \frac{1}{\sqrt{6}}=\frac{2}{\sqrt{6}}=\sqrt{\frac{2}{3}}\)
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Straight Line
88754
If a point ' \(P\) ' on the line \(3 x+5 y=15\) is equidistant from the coordinate axes, then \(P\) lies \(3 x+5 y=15\)
1 Only in the first quadrant
2 Either in first or in second quadrant
3 Either in first or in third quadrant
4 Only in the third quadrant
Explanation:
(B):
\(\frac{x}{5}+\frac{y}{3}=1\)
\(|\mathrm{a}|=|\mathrm{b}|\)
\(|\mathrm{a}|=\left|\frac{15-3 \mathrm{a}}{5}\right|\)
\(\mathrm{a}=\frac{15-3 \mathrm{a}}{5} \quad \mathrm{a}=\frac{-15+3 \mathrm{a}}{5}\)
\(\mathrm{a}=\frac{15}{2} 2 \mathrm{a}=-15\)
\(\mathrm{a}=-15 / 2\)
Either in first or in second quadrant.
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Straight Line
88755
What is the perpendicular distance of a corner to the diagonal not passing through that corner?
1 \(\frac{2}{3}\)
2 \(\frac{\sqrt{3}}{2}\)
3 \(\sqrt{\frac{2}{3}}\)
4 None of the above
Explanation:
(C):
Let the edges \(\mathrm{OA}, \mathrm{OB}, \mathrm{OC}\) of unit cube represents the unit vectors \(\hat{i}, \hat{j}, \hat{k}\) resp. Let \(\mathrm{cm}\) be the, perpendicular from corner on diagonal OP.
We have \(\overrightarrow{\mathrm{OP}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}\)
\(|\overrightarrow{\mathrm{OP}}|=|\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}|=\sqrt{3}\)
Unit vector in the direction of \(\overrightarrow{\mathrm{OP}}\) is \(\frac{\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}}{\sqrt{3}}\)
\(\mathrm{OM}=\) projection of \(\mathrm{OC}\) and \(\mathrm{OP}\)
\(\mathrm{OM}=\overrightarrow{\mathrm{OC}}\) (unit vector along \(\overrightarrow{\mathrm{OP}}\) )
\(O M=\hat{\mathrm{j}} \cdot\left(\frac{\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}}{\sqrt{3}}\right)=\frac{1}{\sqrt{3}}\)
Now, \(\mathrm{CM}^{2}=\mathrm{OC}^{2}-\mathrm{OM}^{2}\)
\(=|\overrightarrow{\mathrm{OC}}|^{2}-\mathrm{OM}^{2}\)
\(\mathrm{CM}^{2}=1-\frac{1}{3}=\frac{2}{3}\)
\(\mathrm{CM}=\sqrt{\frac{2}{3}}\)
