88760
If the straight line passing through the point \(\mathbf{P}(3,4)\) makes an angle \(\frac{\pi}{6}\) with the positive direction of \(X\)-axis and meets the line \(12 x+5 y\) \(+10=0\) at \(Q\), then the length of \(P Q\) is
1 \(\frac{44}{12 \sqrt{2}+9}\)
2 \(\frac{66}{12 \sqrt{3}+5}\)
3 \(\frac{132}{12 \sqrt{3}+5}\)
4 \(\frac{148}{6 \sqrt{2}+3}\)
Explanation:
(C) : The equation of line passes through point \(\mathrm{P}(3,4)\) makes an angle \(\frac{\pi}{4}\) with the positive direction of X-axis is \(\frac{x-3}{\cos \frac{\pi}{6}}=\frac{y-4}{\sin \frac{\pi}{6}}=r\)
So, let point \(A\left(3+\frac{\sqrt{3} r}{2}, 4+\frac{r}{2}\right)\), since, the point \(Q\) on the line \(12 x+5 y+10=0\)
So, \(\quad 36+6 \sqrt{3} \mathrm{r}+20+\frac{5}{2} \mathrm{r}+10=0\)
\(\Rightarrow 132+(12 \sqrt{3}+5) \mathrm{r}=0 \Rightarrow \mathrm{r}=-\frac{132}{12 \sqrt{3}+5}\)
\(\therefore\) The length PQ is \(|\mathrm{r}|=\frac{132}{12 \sqrt{3}+5}\)
TS EAMCET-2020-11.09.2020
Straight Line
88761
Let \(Q\) be the foot of perpendicular drawn from the point \(P(1,2,3)\) to the plane \(x+2 y+z=14\). If \(R\) is a point on the plane such that \(\angle P R Q=\) \(60^{\circ}\), then the area of \(\triangle \mathrm{PQR}\) is equal to:
1 \(\frac{\sqrt{3}}{2}\)
2 \(\sqrt{3}\)
3 \(2 \sqrt{3}\)
4 3
Explanation:
(B) :
Length of perpendicular
\(P Q =\left|\frac{1+4+3-14}{\sqrt{6}}\right|=\sqrt{6}\)
\(Q R =(P Q) \cot 60^{\circ}=\sqrt{2}\)
\(\therefore\) Area of \(\Delta \mathrm{PQR}=\frac{1}{2}(\mathrm{PQ})(\mathrm{QR})=\sqrt{3}\)
JEE Main-2022-29.07.2022
Straight Line
88762
The distance between the lines \(3 x+4 y=9\) and \(6 x+8 y=15\) is
1 \(0 \cdot 3\) units
2 5 units
3 0.5 units
4 3 units
Explanation:
(A) : We have given this line
\(3 x+4 y=9\)
and \(\quad 6 x+8 y=15\)
\(2(3 x+4 y)=15\)
\(3 x+4 y=\frac{15}{2}\)
Euqation (i) and equation (ii) one parallel
\(\mathrm{d}=\left|\frac{9 \frac{15}{2}}{\sqrt{9+16}}\right|=\left|\frac{\frac{18-15}{2}}{\sqrt{25}}\right|\)
\(\mathrm{d}=\frac{3}{2 \times 1}=\frac{3}{10}=0.3\)
MHT CET-2021
Straight Line
88763
The X-intercept of a line passing through the points \(\left(\frac{-1}{2}, 1\right)\) and \((1,2)\) is
1 1
2 -2
3 -1
4 3
Explanation:
(B) :
The equation of line through the points
\(\left(\frac{-1}{2}, 1\right)\) and \((1,2)\) is
\((\mathrm{y}-1)=\frac{2-1}{1+\frac{1}{2}}\left(\mathrm{x}+\frac{1}{2}\right)\)
Or \(y-1=\frac{1}{3 / 2}\left(x+\frac{1}{2}\right)=\frac{2}{3} x+\frac{1}{3}\)
Put \(\mathrm{y}=0\) to \(\mathrm{x}\)-intercept.
\(0-1=\frac{2}{3} x+\frac{1}{3} \Rightarrow \frac{2}{3} x=-1-\frac{1}{3}=\frac{-4}{3}\)
\(\frac{2 x}{3}=\frac{-4}{3}\)
\(\mathrm{x}=-2\)
88760
If the straight line passing through the point \(\mathbf{P}(3,4)\) makes an angle \(\frac{\pi}{6}\) with the positive direction of \(X\)-axis and meets the line \(12 x+5 y\) \(+10=0\) at \(Q\), then the length of \(P Q\) is
1 \(\frac{44}{12 \sqrt{2}+9}\)
2 \(\frac{66}{12 \sqrt{3}+5}\)
3 \(\frac{132}{12 \sqrt{3}+5}\)
4 \(\frac{148}{6 \sqrt{2}+3}\)
Explanation:
(C) : The equation of line passes through point \(\mathrm{P}(3,4)\) makes an angle \(\frac{\pi}{4}\) with the positive direction of X-axis is \(\frac{x-3}{\cos \frac{\pi}{6}}=\frac{y-4}{\sin \frac{\pi}{6}}=r\)
So, let point \(A\left(3+\frac{\sqrt{3} r}{2}, 4+\frac{r}{2}\right)\), since, the point \(Q\) on the line \(12 x+5 y+10=0\)
So, \(\quad 36+6 \sqrt{3} \mathrm{r}+20+\frac{5}{2} \mathrm{r}+10=0\)
\(\Rightarrow 132+(12 \sqrt{3}+5) \mathrm{r}=0 \Rightarrow \mathrm{r}=-\frac{132}{12 \sqrt{3}+5}\)
\(\therefore\) The length PQ is \(|\mathrm{r}|=\frac{132}{12 \sqrt{3}+5}\)
TS EAMCET-2020-11.09.2020
Straight Line
88761
Let \(Q\) be the foot of perpendicular drawn from the point \(P(1,2,3)\) to the plane \(x+2 y+z=14\). If \(R\) is a point on the plane such that \(\angle P R Q=\) \(60^{\circ}\), then the area of \(\triangle \mathrm{PQR}\) is equal to:
1 \(\frac{\sqrt{3}}{2}\)
2 \(\sqrt{3}\)
3 \(2 \sqrt{3}\)
4 3
Explanation:
(B) :
Length of perpendicular
\(P Q =\left|\frac{1+4+3-14}{\sqrt{6}}\right|=\sqrt{6}\)
\(Q R =(P Q) \cot 60^{\circ}=\sqrt{2}\)
\(\therefore\) Area of \(\Delta \mathrm{PQR}=\frac{1}{2}(\mathrm{PQ})(\mathrm{QR})=\sqrt{3}\)
JEE Main-2022-29.07.2022
Straight Line
88762
The distance between the lines \(3 x+4 y=9\) and \(6 x+8 y=15\) is
1 \(0 \cdot 3\) units
2 5 units
3 0.5 units
4 3 units
Explanation:
(A) : We have given this line
\(3 x+4 y=9\)
and \(\quad 6 x+8 y=15\)
\(2(3 x+4 y)=15\)
\(3 x+4 y=\frac{15}{2}\)
Euqation (i) and equation (ii) one parallel
\(\mathrm{d}=\left|\frac{9 \frac{15}{2}}{\sqrt{9+16}}\right|=\left|\frac{\frac{18-15}{2}}{\sqrt{25}}\right|\)
\(\mathrm{d}=\frac{3}{2 \times 1}=\frac{3}{10}=0.3\)
MHT CET-2021
Straight Line
88763
The X-intercept of a line passing through the points \(\left(\frac{-1}{2}, 1\right)\) and \((1,2)\) is
1 1
2 -2
3 -1
4 3
Explanation:
(B) :
The equation of line through the points
\(\left(\frac{-1}{2}, 1\right)\) and \((1,2)\) is
\((\mathrm{y}-1)=\frac{2-1}{1+\frac{1}{2}}\left(\mathrm{x}+\frac{1}{2}\right)\)
Or \(y-1=\frac{1}{3 / 2}\left(x+\frac{1}{2}\right)=\frac{2}{3} x+\frac{1}{3}\)
Put \(\mathrm{y}=0\) to \(\mathrm{x}\)-intercept.
\(0-1=\frac{2}{3} x+\frac{1}{3} \Rightarrow \frac{2}{3} x=-1-\frac{1}{3}=\frac{-4}{3}\)
\(\frac{2 x}{3}=\frac{-4}{3}\)
\(\mathrm{x}=-2\)
NEET Test Series from KOTA - 10 Papers In MS WORD
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Straight Line
88760
If the straight line passing through the point \(\mathbf{P}(3,4)\) makes an angle \(\frac{\pi}{6}\) with the positive direction of \(X\)-axis and meets the line \(12 x+5 y\) \(+10=0\) at \(Q\), then the length of \(P Q\) is
1 \(\frac{44}{12 \sqrt{2}+9}\)
2 \(\frac{66}{12 \sqrt{3}+5}\)
3 \(\frac{132}{12 \sqrt{3}+5}\)
4 \(\frac{148}{6 \sqrt{2}+3}\)
Explanation:
(C) : The equation of line passes through point \(\mathrm{P}(3,4)\) makes an angle \(\frac{\pi}{4}\) with the positive direction of X-axis is \(\frac{x-3}{\cos \frac{\pi}{6}}=\frac{y-4}{\sin \frac{\pi}{6}}=r\)
So, let point \(A\left(3+\frac{\sqrt{3} r}{2}, 4+\frac{r}{2}\right)\), since, the point \(Q\) on the line \(12 x+5 y+10=0\)
So, \(\quad 36+6 \sqrt{3} \mathrm{r}+20+\frac{5}{2} \mathrm{r}+10=0\)
\(\Rightarrow 132+(12 \sqrt{3}+5) \mathrm{r}=0 \Rightarrow \mathrm{r}=-\frac{132}{12 \sqrt{3}+5}\)
\(\therefore\) The length PQ is \(|\mathrm{r}|=\frac{132}{12 \sqrt{3}+5}\)
TS EAMCET-2020-11.09.2020
Straight Line
88761
Let \(Q\) be the foot of perpendicular drawn from the point \(P(1,2,3)\) to the plane \(x+2 y+z=14\). If \(R\) is a point on the plane such that \(\angle P R Q=\) \(60^{\circ}\), then the area of \(\triangle \mathrm{PQR}\) is equal to:
1 \(\frac{\sqrt{3}}{2}\)
2 \(\sqrt{3}\)
3 \(2 \sqrt{3}\)
4 3
Explanation:
(B) :
Length of perpendicular
\(P Q =\left|\frac{1+4+3-14}{\sqrt{6}}\right|=\sqrt{6}\)
\(Q R =(P Q) \cot 60^{\circ}=\sqrt{2}\)
\(\therefore\) Area of \(\Delta \mathrm{PQR}=\frac{1}{2}(\mathrm{PQ})(\mathrm{QR})=\sqrt{3}\)
JEE Main-2022-29.07.2022
Straight Line
88762
The distance between the lines \(3 x+4 y=9\) and \(6 x+8 y=15\) is
1 \(0 \cdot 3\) units
2 5 units
3 0.5 units
4 3 units
Explanation:
(A) : We have given this line
\(3 x+4 y=9\)
and \(\quad 6 x+8 y=15\)
\(2(3 x+4 y)=15\)
\(3 x+4 y=\frac{15}{2}\)
Euqation (i) and equation (ii) one parallel
\(\mathrm{d}=\left|\frac{9 \frac{15}{2}}{\sqrt{9+16}}\right|=\left|\frac{\frac{18-15}{2}}{\sqrt{25}}\right|\)
\(\mathrm{d}=\frac{3}{2 \times 1}=\frac{3}{10}=0.3\)
MHT CET-2021
Straight Line
88763
The X-intercept of a line passing through the points \(\left(\frac{-1}{2}, 1\right)\) and \((1,2)\) is
1 1
2 -2
3 -1
4 3
Explanation:
(B) :
The equation of line through the points
\(\left(\frac{-1}{2}, 1\right)\) and \((1,2)\) is
\((\mathrm{y}-1)=\frac{2-1}{1+\frac{1}{2}}\left(\mathrm{x}+\frac{1}{2}\right)\)
Or \(y-1=\frac{1}{3 / 2}\left(x+\frac{1}{2}\right)=\frac{2}{3} x+\frac{1}{3}\)
Put \(\mathrm{y}=0\) to \(\mathrm{x}\)-intercept.
\(0-1=\frac{2}{3} x+\frac{1}{3} \Rightarrow \frac{2}{3} x=-1-\frac{1}{3}=\frac{-4}{3}\)
\(\frac{2 x}{3}=\frac{-4}{3}\)
\(\mathrm{x}=-2\)
88760
If the straight line passing through the point \(\mathbf{P}(3,4)\) makes an angle \(\frac{\pi}{6}\) with the positive direction of \(X\)-axis and meets the line \(12 x+5 y\) \(+10=0\) at \(Q\), then the length of \(P Q\) is
1 \(\frac{44}{12 \sqrt{2}+9}\)
2 \(\frac{66}{12 \sqrt{3}+5}\)
3 \(\frac{132}{12 \sqrt{3}+5}\)
4 \(\frac{148}{6 \sqrt{2}+3}\)
Explanation:
(C) : The equation of line passes through point \(\mathrm{P}(3,4)\) makes an angle \(\frac{\pi}{4}\) with the positive direction of X-axis is \(\frac{x-3}{\cos \frac{\pi}{6}}=\frac{y-4}{\sin \frac{\pi}{6}}=r\)
So, let point \(A\left(3+\frac{\sqrt{3} r}{2}, 4+\frac{r}{2}\right)\), since, the point \(Q\) on the line \(12 x+5 y+10=0\)
So, \(\quad 36+6 \sqrt{3} \mathrm{r}+20+\frac{5}{2} \mathrm{r}+10=0\)
\(\Rightarrow 132+(12 \sqrt{3}+5) \mathrm{r}=0 \Rightarrow \mathrm{r}=-\frac{132}{12 \sqrt{3}+5}\)
\(\therefore\) The length PQ is \(|\mathrm{r}|=\frac{132}{12 \sqrt{3}+5}\)
TS EAMCET-2020-11.09.2020
Straight Line
88761
Let \(Q\) be the foot of perpendicular drawn from the point \(P(1,2,3)\) to the plane \(x+2 y+z=14\). If \(R\) is a point on the plane such that \(\angle P R Q=\) \(60^{\circ}\), then the area of \(\triangle \mathrm{PQR}\) is equal to:
1 \(\frac{\sqrt{3}}{2}\)
2 \(\sqrt{3}\)
3 \(2 \sqrt{3}\)
4 3
Explanation:
(B) :
Length of perpendicular
\(P Q =\left|\frac{1+4+3-14}{\sqrt{6}}\right|=\sqrt{6}\)
\(Q R =(P Q) \cot 60^{\circ}=\sqrt{2}\)
\(\therefore\) Area of \(\Delta \mathrm{PQR}=\frac{1}{2}(\mathrm{PQ})(\mathrm{QR})=\sqrt{3}\)
JEE Main-2022-29.07.2022
Straight Line
88762
The distance between the lines \(3 x+4 y=9\) and \(6 x+8 y=15\) is
1 \(0 \cdot 3\) units
2 5 units
3 0.5 units
4 3 units
Explanation:
(A) : We have given this line
\(3 x+4 y=9\)
and \(\quad 6 x+8 y=15\)
\(2(3 x+4 y)=15\)
\(3 x+4 y=\frac{15}{2}\)
Euqation (i) and equation (ii) one parallel
\(\mathrm{d}=\left|\frac{9 \frac{15}{2}}{\sqrt{9+16}}\right|=\left|\frac{\frac{18-15}{2}}{\sqrt{25}}\right|\)
\(\mathrm{d}=\frac{3}{2 \times 1}=\frac{3}{10}=0.3\)
MHT CET-2021
Straight Line
88763
The X-intercept of a line passing through the points \(\left(\frac{-1}{2}, 1\right)\) and \((1,2)\) is
1 1
2 -2
3 -1
4 3
Explanation:
(B) :
The equation of line through the points
\(\left(\frac{-1}{2}, 1\right)\) and \((1,2)\) is
\((\mathrm{y}-1)=\frac{2-1}{1+\frac{1}{2}}\left(\mathrm{x}+\frac{1}{2}\right)\)
Or \(y-1=\frac{1}{3 / 2}\left(x+\frac{1}{2}\right)=\frac{2}{3} x+\frac{1}{3}\)
Put \(\mathrm{y}=0\) to \(\mathrm{x}\)-intercept.
\(0-1=\frac{2}{3} x+\frac{1}{3} \Rightarrow \frac{2}{3} x=-1-\frac{1}{3}=\frac{-4}{3}\)
\(\frac{2 x}{3}=\frac{-4}{3}\)
\(\mathrm{x}=-2\)
