NEET Test Series from KOTA - 10 Papers In MS WORD
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Straight Line
88747
If \(p\) is the length of perpendicular from origin to the line whose intercepts on the axes are \(a\) and \(b\), then \(\frac{1}{a^{2}}+\frac{1}{b^{2}}\) is equal to
1 \(\mathrm{p}^{2}\)
2 \(\frac{1}{p^{2}}\)
3 \(2 \mathrm{p}^{2}\)
4 \(\frac{1}{2 p^{2}}\)
Explanation:
(B) : Equation of line in intercept form is
\(\frac{x}{a}+\frac{y}{b}=1 \Rightarrow \frac{x}{a}+\frac{y}{b}-1=0\)
Its distance from origin is \(\mathrm{p}\)
\(\therefore \quad\left|\frac{0+0-1}{\sqrt{\frac{1}{\mathrm{a}^{2}}+\frac{1}{\mathrm{~b}^{2}}}}\right|=\mathrm{p}\)
\(\Rightarrow p=\frac{1}{\sqrt{\frac{1}{a^{2}}+\frac{1}{b^{2}}}}\)
\(\Rightarrow \frac{1}{\mathrm{p}}=\sqrt{\frac{1}{\mathrm{a}^{2}}+\frac{1}{\mathrm{~b}^{2}}}\)
Now, squaring both sides, we get \(\frac{1}{\mathrm{p}^{2}}=\frac{1}{\mathrm{a}^{2}}+\frac{1}{\mathrm{~b}^{2}}\)
COMEDK-2019
Straight Line
88748
The point on the line \(4 x+3 y=5\), which is equidistant from \((1,2)\) and \((3,4)\), is
1 \((7,-4)\)
2 \((-10,15)\)
3 \(\left(\frac{1}{7}, \frac{8}{7}\right)\)
4 \(\left(0, \frac{5}{4}\right)\)
Explanation:
(B) : Let the point \((x, y)\) be on the line \(4 x+3 y=5\) \(\therefore \quad 4 \mathrm{x}_{1}+3 \mathrm{y}_{1}=5\)
Distance of point \(\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)\) from \((1,2)\)
\(\therefore \quad\left(\mathrm{x}_{1}-1\right)^{2}+\left(\mathrm{y}_{1}-2\right)^{2}\)
And,
Distance of point \(\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)\) from \((3,4)\)
\(\therefore \quad\left(\mathrm{x}_{1}-3\right)^{2}+\left(\mathrm{y}_{1}-4\right)^{2}\)
Given that the point \(\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)\) is equidistant from \((1,2)\) and \((3,4)\)
\(\therefore\left(\mathrm{x}_{1}-1\right)^{2}+\left(\mathrm{y}_{1}-2\right)^{2}=\left(\mathrm{x}_{1}-3\right)^{2}+\left(\mathrm{y}_{1}-4\right)^{2}\)
\(\mathrm{x}_{1}{ }^{2}+1-2 \mathrm{x},+\mathrm{y}^{2}+4-4 \mathrm{y}_{1}=\mathrm{x}_{1}{ }^{2}+9-6 \mathrm{x}_{1}+\mathrm{y}_{1}{ }^{2}+16-\)
\(8 y\)
\(4 x_{1}+4 y_{1}=20\)
\(x_{1}+y_{1}=5 \tag{ii}\)
Form equation (i) and (ii) we get,
\(\mathrm{y}_{1}=15 \text { and } \quad \mathrm{x}_{1}=-10\)
Hence, the point will be \((-10,15)\)
UPSEE-2015
Straight Line
88749
Let \(P Q R\) be a right angled isosceles triangle, right angled at \(P(2,1)\). If the equation of the line \(Q R\) is \(2 x+y=3\), Then the equation representing the pair of lines \(P Q\) and \(P R\) is
1 \(3 x^{2}-3 y^{2}+8 x y+20 x+10 y+25=0\)
2 \(3 x^{2}-3 y^{2}+8 x y-20 x-10 y+25=0\)
3 \(3 x^{2}-3 y^{2}+8 x y+10 x+15 y+20=0\)
4 \(3 x^{2}-3 y^{2}-8 x y-10 x-15 y-20=0\)
Explanation:
(B) : The equation of PQ and PR are given by,
\(y-1=\frac{-2 \pm \tan 45^{\circ}}{1 \pm\left(-2 \tan 45^{\circ}\right)}(x-2)\)
\(\mathrm{y}-1=\left(\frac{-2 \mp 1}{1 \pm 2}\right)(\mathrm{x}-2)\)
\(\mathrm{y}-1=-\frac{1}{3}(\mathrm{x}-2)\)
and \(\quad \mathrm{y}-1=3(\mathrm{x}-2)\)
\(x+3 y=5\) and \(3 x-y=5\)
The combined equation of two line is,
\((x+3 y-5)(3 x-y-5)=0\)
Hence, \(3 x^{2}-3 y^{2}+8 x y-20 x-10 y+25=0\)
UPSEE-2010
Straight Line
88750
The distance of the point \(A(a, b, c)\) from the \(X\) axis is
1 a
2 \(\sqrt{b^{2}+c^{2}}\)
3 \(\sqrt{a^{2}+b^{2}}\)
4 \(a^{2}+b^{2}\)
Explanation:
(B) : Given the distance between the point
\(\mathrm{P}(\mathrm{a}, \mathrm{b}, \mathrm{c})\) and \(\mathrm{x}\)-axis
Distance \((d)=\sqrt{(a-a)^{2}+(0-b)^{2}+(0-c)^{2}}\)
\(\mathrm{d}=\sqrt{0^{2}+(-\mathrm{b})^{2}+(-\mathrm{c})^{2}}\)
\(\mathrm{d}=\sqrt{\mathrm{b}^{2}+\mathrm{c}^{2}}\)
88747
If \(p\) is the length of perpendicular from origin to the line whose intercepts on the axes are \(a\) and \(b\), then \(\frac{1}{a^{2}}+\frac{1}{b^{2}}\) is equal to
1 \(\mathrm{p}^{2}\)
2 \(\frac{1}{p^{2}}\)
3 \(2 \mathrm{p}^{2}\)
4 \(\frac{1}{2 p^{2}}\)
Explanation:
(B) : Equation of line in intercept form is
\(\frac{x}{a}+\frac{y}{b}=1 \Rightarrow \frac{x}{a}+\frac{y}{b}-1=0\)
Its distance from origin is \(\mathrm{p}\)
\(\therefore \quad\left|\frac{0+0-1}{\sqrt{\frac{1}{\mathrm{a}^{2}}+\frac{1}{\mathrm{~b}^{2}}}}\right|=\mathrm{p}\)
\(\Rightarrow p=\frac{1}{\sqrt{\frac{1}{a^{2}}+\frac{1}{b^{2}}}}\)
\(\Rightarrow \frac{1}{\mathrm{p}}=\sqrt{\frac{1}{\mathrm{a}^{2}}+\frac{1}{\mathrm{~b}^{2}}}\)
Now, squaring both sides, we get \(\frac{1}{\mathrm{p}^{2}}=\frac{1}{\mathrm{a}^{2}}+\frac{1}{\mathrm{~b}^{2}}\)
COMEDK-2019
Straight Line
88748
The point on the line \(4 x+3 y=5\), which is equidistant from \((1,2)\) and \((3,4)\), is
1 \((7,-4)\)
2 \((-10,15)\)
3 \(\left(\frac{1}{7}, \frac{8}{7}\right)\)
4 \(\left(0, \frac{5}{4}\right)\)
Explanation:
(B) : Let the point \((x, y)\) be on the line \(4 x+3 y=5\) \(\therefore \quad 4 \mathrm{x}_{1}+3 \mathrm{y}_{1}=5\)
Distance of point \(\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)\) from \((1,2)\)
\(\therefore \quad\left(\mathrm{x}_{1}-1\right)^{2}+\left(\mathrm{y}_{1}-2\right)^{2}\)
And,
Distance of point \(\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)\) from \((3,4)\)
\(\therefore \quad\left(\mathrm{x}_{1}-3\right)^{2}+\left(\mathrm{y}_{1}-4\right)^{2}\)
Given that the point \(\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)\) is equidistant from \((1,2)\) and \((3,4)\)
\(\therefore\left(\mathrm{x}_{1}-1\right)^{2}+\left(\mathrm{y}_{1}-2\right)^{2}=\left(\mathrm{x}_{1}-3\right)^{2}+\left(\mathrm{y}_{1}-4\right)^{2}\)
\(\mathrm{x}_{1}{ }^{2}+1-2 \mathrm{x},+\mathrm{y}^{2}+4-4 \mathrm{y}_{1}=\mathrm{x}_{1}{ }^{2}+9-6 \mathrm{x}_{1}+\mathrm{y}_{1}{ }^{2}+16-\)
\(8 y\)
\(4 x_{1}+4 y_{1}=20\)
\(x_{1}+y_{1}=5 \tag{ii}\)
Form equation (i) and (ii) we get,
\(\mathrm{y}_{1}=15 \text { and } \quad \mathrm{x}_{1}=-10\)
Hence, the point will be \((-10,15)\)
UPSEE-2015
Straight Line
88749
Let \(P Q R\) be a right angled isosceles triangle, right angled at \(P(2,1)\). If the equation of the line \(Q R\) is \(2 x+y=3\), Then the equation representing the pair of lines \(P Q\) and \(P R\) is
1 \(3 x^{2}-3 y^{2}+8 x y+20 x+10 y+25=0\)
2 \(3 x^{2}-3 y^{2}+8 x y-20 x-10 y+25=0\)
3 \(3 x^{2}-3 y^{2}+8 x y+10 x+15 y+20=0\)
4 \(3 x^{2}-3 y^{2}-8 x y-10 x-15 y-20=0\)
Explanation:
(B) : The equation of PQ and PR are given by,
\(y-1=\frac{-2 \pm \tan 45^{\circ}}{1 \pm\left(-2 \tan 45^{\circ}\right)}(x-2)\)
\(\mathrm{y}-1=\left(\frac{-2 \mp 1}{1 \pm 2}\right)(\mathrm{x}-2)\)
\(\mathrm{y}-1=-\frac{1}{3}(\mathrm{x}-2)\)
and \(\quad \mathrm{y}-1=3(\mathrm{x}-2)\)
\(x+3 y=5\) and \(3 x-y=5\)
The combined equation of two line is,
\((x+3 y-5)(3 x-y-5)=0\)
Hence, \(3 x^{2}-3 y^{2}+8 x y-20 x-10 y+25=0\)
UPSEE-2010
Straight Line
88750
The distance of the point \(A(a, b, c)\) from the \(X\) axis is
1 a
2 \(\sqrt{b^{2}+c^{2}}\)
3 \(\sqrt{a^{2}+b^{2}}\)
4 \(a^{2}+b^{2}\)
Explanation:
(B) : Given the distance between the point
\(\mathrm{P}(\mathrm{a}, \mathrm{b}, \mathrm{c})\) and \(\mathrm{x}\)-axis
Distance \((d)=\sqrt{(a-a)^{2}+(0-b)^{2}+(0-c)^{2}}\)
\(\mathrm{d}=\sqrt{0^{2}+(-\mathrm{b})^{2}+(-\mathrm{c})^{2}}\)
\(\mathrm{d}=\sqrt{\mathrm{b}^{2}+\mathrm{c}^{2}}\)
88747
If \(p\) is the length of perpendicular from origin to the line whose intercepts on the axes are \(a\) and \(b\), then \(\frac{1}{a^{2}}+\frac{1}{b^{2}}\) is equal to
1 \(\mathrm{p}^{2}\)
2 \(\frac{1}{p^{2}}\)
3 \(2 \mathrm{p}^{2}\)
4 \(\frac{1}{2 p^{2}}\)
Explanation:
(B) : Equation of line in intercept form is
\(\frac{x}{a}+\frac{y}{b}=1 \Rightarrow \frac{x}{a}+\frac{y}{b}-1=0\)
Its distance from origin is \(\mathrm{p}\)
\(\therefore \quad\left|\frac{0+0-1}{\sqrt{\frac{1}{\mathrm{a}^{2}}+\frac{1}{\mathrm{~b}^{2}}}}\right|=\mathrm{p}\)
\(\Rightarrow p=\frac{1}{\sqrt{\frac{1}{a^{2}}+\frac{1}{b^{2}}}}\)
\(\Rightarrow \frac{1}{\mathrm{p}}=\sqrt{\frac{1}{\mathrm{a}^{2}}+\frac{1}{\mathrm{~b}^{2}}}\)
Now, squaring both sides, we get \(\frac{1}{\mathrm{p}^{2}}=\frac{1}{\mathrm{a}^{2}}+\frac{1}{\mathrm{~b}^{2}}\)
COMEDK-2019
Straight Line
88748
The point on the line \(4 x+3 y=5\), which is equidistant from \((1,2)\) and \((3,4)\), is
1 \((7,-4)\)
2 \((-10,15)\)
3 \(\left(\frac{1}{7}, \frac{8}{7}\right)\)
4 \(\left(0, \frac{5}{4}\right)\)
Explanation:
(B) : Let the point \((x, y)\) be on the line \(4 x+3 y=5\) \(\therefore \quad 4 \mathrm{x}_{1}+3 \mathrm{y}_{1}=5\)
Distance of point \(\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)\) from \((1,2)\)
\(\therefore \quad\left(\mathrm{x}_{1}-1\right)^{2}+\left(\mathrm{y}_{1}-2\right)^{2}\)
And,
Distance of point \(\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)\) from \((3,4)\)
\(\therefore \quad\left(\mathrm{x}_{1}-3\right)^{2}+\left(\mathrm{y}_{1}-4\right)^{2}\)
Given that the point \(\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)\) is equidistant from \((1,2)\) and \((3,4)\)
\(\therefore\left(\mathrm{x}_{1}-1\right)^{2}+\left(\mathrm{y}_{1}-2\right)^{2}=\left(\mathrm{x}_{1}-3\right)^{2}+\left(\mathrm{y}_{1}-4\right)^{2}\)
\(\mathrm{x}_{1}{ }^{2}+1-2 \mathrm{x},+\mathrm{y}^{2}+4-4 \mathrm{y}_{1}=\mathrm{x}_{1}{ }^{2}+9-6 \mathrm{x}_{1}+\mathrm{y}_{1}{ }^{2}+16-\)
\(8 y\)
\(4 x_{1}+4 y_{1}=20\)
\(x_{1}+y_{1}=5 \tag{ii}\)
Form equation (i) and (ii) we get,
\(\mathrm{y}_{1}=15 \text { and } \quad \mathrm{x}_{1}=-10\)
Hence, the point will be \((-10,15)\)
UPSEE-2015
Straight Line
88749
Let \(P Q R\) be a right angled isosceles triangle, right angled at \(P(2,1)\). If the equation of the line \(Q R\) is \(2 x+y=3\), Then the equation representing the pair of lines \(P Q\) and \(P R\) is
1 \(3 x^{2}-3 y^{2}+8 x y+20 x+10 y+25=0\)
2 \(3 x^{2}-3 y^{2}+8 x y-20 x-10 y+25=0\)
3 \(3 x^{2}-3 y^{2}+8 x y+10 x+15 y+20=0\)
4 \(3 x^{2}-3 y^{2}-8 x y-10 x-15 y-20=0\)
Explanation:
(B) : The equation of PQ and PR are given by,
\(y-1=\frac{-2 \pm \tan 45^{\circ}}{1 \pm\left(-2 \tan 45^{\circ}\right)}(x-2)\)
\(\mathrm{y}-1=\left(\frac{-2 \mp 1}{1 \pm 2}\right)(\mathrm{x}-2)\)
\(\mathrm{y}-1=-\frac{1}{3}(\mathrm{x}-2)\)
and \(\quad \mathrm{y}-1=3(\mathrm{x}-2)\)
\(x+3 y=5\) and \(3 x-y=5\)
The combined equation of two line is,
\((x+3 y-5)(3 x-y-5)=0\)
Hence, \(3 x^{2}-3 y^{2}+8 x y-20 x-10 y+25=0\)
UPSEE-2010
Straight Line
88750
The distance of the point \(A(a, b, c)\) from the \(X\) axis is
1 a
2 \(\sqrt{b^{2}+c^{2}}\)
3 \(\sqrt{a^{2}+b^{2}}\)
4 \(a^{2}+b^{2}\)
Explanation:
(B) : Given the distance between the point
\(\mathrm{P}(\mathrm{a}, \mathrm{b}, \mathrm{c})\) and \(\mathrm{x}\)-axis
Distance \((d)=\sqrt{(a-a)^{2}+(0-b)^{2}+(0-c)^{2}}\)
\(\mathrm{d}=\sqrt{0^{2}+(-\mathrm{b})^{2}+(-\mathrm{c})^{2}}\)
\(\mathrm{d}=\sqrt{\mathrm{b}^{2}+\mathrm{c}^{2}}\)
88747
If \(p\) is the length of perpendicular from origin to the line whose intercepts on the axes are \(a\) and \(b\), then \(\frac{1}{a^{2}}+\frac{1}{b^{2}}\) is equal to
1 \(\mathrm{p}^{2}\)
2 \(\frac{1}{p^{2}}\)
3 \(2 \mathrm{p}^{2}\)
4 \(\frac{1}{2 p^{2}}\)
Explanation:
(B) : Equation of line in intercept form is
\(\frac{x}{a}+\frac{y}{b}=1 \Rightarrow \frac{x}{a}+\frac{y}{b}-1=0\)
Its distance from origin is \(\mathrm{p}\)
\(\therefore \quad\left|\frac{0+0-1}{\sqrt{\frac{1}{\mathrm{a}^{2}}+\frac{1}{\mathrm{~b}^{2}}}}\right|=\mathrm{p}\)
\(\Rightarrow p=\frac{1}{\sqrt{\frac{1}{a^{2}}+\frac{1}{b^{2}}}}\)
\(\Rightarrow \frac{1}{\mathrm{p}}=\sqrt{\frac{1}{\mathrm{a}^{2}}+\frac{1}{\mathrm{~b}^{2}}}\)
Now, squaring both sides, we get \(\frac{1}{\mathrm{p}^{2}}=\frac{1}{\mathrm{a}^{2}}+\frac{1}{\mathrm{~b}^{2}}\)
COMEDK-2019
Straight Line
88748
The point on the line \(4 x+3 y=5\), which is equidistant from \((1,2)\) and \((3,4)\), is
1 \((7,-4)\)
2 \((-10,15)\)
3 \(\left(\frac{1}{7}, \frac{8}{7}\right)\)
4 \(\left(0, \frac{5}{4}\right)\)
Explanation:
(B) : Let the point \((x, y)\) be on the line \(4 x+3 y=5\) \(\therefore \quad 4 \mathrm{x}_{1}+3 \mathrm{y}_{1}=5\)
Distance of point \(\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)\) from \((1,2)\)
\(\therefore \quad\left(\mathrm{x}_{1}-1\right)^{2}+\left(\mathrm{y}_{1}-2\right)^{2}\)
And,
Distance of point \(\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)\) from \((3,4)\)
\(\therefore \quad\left(\mathrm{x}_{1}-3\right)^{2}+\left(\mathrm{y}_{1}-4\right)^{2}\)
Given that the point \(\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)\) is equidistant from \((1,2)\) and \((3,4)\)
\(\therefore\left(\mathrm{x}_{1}-1\right)^{2}+\left(\mathrm{y}_{1}-2\right)^{2}=\left(\mathrm{x}_{1}-3\right)^{2}+\left(\mathrm{y}_{1}-4\right)^{2}\)
\(\mathrm{x}_{1}{ }^{2}+1-2 \mathrm{x},+\mathrm{y}^{2}+4-4 \mathrm{y}_{1}=\mathrm{x}_{1}{ }^{2}+9-6 \mathrm{x}_{1}+\mathrm{y}_{1}{ }^{2}+16-\)
\(8 y\)
\(4 x_{1}+4 y_{1}=20\)
\(x_{1}+y_{1}=5 \tag{ii}\)
Form equation (i) and (ii) we get,
\(\mathrm{y}_{1}=15 \text { and } \quad \mathrm{x}_{1}=-10\)
Hence, the point will be \((-10,15)\)
UPSEE-2015
Straight Line
88749
Let \(P Q R\) be a right angled isosceles triangle, right angled at \(P(2,1)\). If the equation of the line \(Q R\) is \(2 x+y=3\), Then the equation representing the pair of lines \(P Q\) and \(P R\) is
1 \(3 x^{2}-3 y^{2}+8 x y+20 x+10 y+25=0\)
2 \(3 x^{2}-3 y^{2}+8 x y-20 x-10 y+25=0\)
3 \(3 x^{2}-3 y^{2}+8 x y+10 x+15 y+20=0\)
4 \(3 x^{2}-3 y^{2}-8 x y-10 x-15 y-20=0\)
Explanation:
(B) : The equation of PQ and PR are given by,
\(y-1=\frac{-2 \pm \tan 45^{\circ}}{1 \pm\left(-2 \tan 45^{\circ}\right)}(x-2)\)
\(\mathrm{y}-1=\left(\frac{-2 \mp 1}{1 \pm 2}\right)(\mathrm{x}-2)\)
\(\mathrm{y}-1=-\frac{1}{3}(\mathrm{x}-2)\)
and \(\quad \mathrm{y}-1=3(\mathrm{x}-2)\)
\(x+3 y=5\) and \(3 x-y=5\)
The combined equation of two line is,
\((x+3 y-5)(3 x-y-5)=0\)
Hence, \(3 x^{2}-3 y^{2}+8 x y-20 x-10 y+25=0\)
UPSEE-2010
Straight Line
88750
The distance of the point \(A(a, b, c)\) from the \(X\) axis is
1 a
2 \(\sqrt{b^{2}+c^{2}}\)
3 \(\sqrt{a^{2}+b^{2}}\)
4 \(a^{2}+b^{2}\)
Explanation:
(B) : Given the distance between the point
\(\mathrm{P}(\mathrm{a}, \mathrm{b}, \mathrm{c})\) and \(\mathrm{x}\)-axis
Distance \((d)=\sqrt{(a-a)^{2}+(0-b)^{2}+(0-c)^{2}}\)
\(\mathrm{d}=\sqrt{0^{2}+(-\mathrm{b})^{2}+(-\mathrm{c})^{2}}\)
\(\mathrm{d}=\sqrt{\mathrm{b}^{2}+\mathrm{c}^{2}}\)
