88263
If \(a, c, b\) are in G.P., then the area of the triangle formed by the lines \(a x+b y+c=0\) with the coordinates axes is equal to
1 1
2 2
3 \(\frac{1}{2}\)
4 Nones of these
Explanation:
(C) : Areas of the triangle \(=\frac{1}{2}(\mathrm{x}\) intercept \() \times(\mathrm{y}\) intercept \()\) \(=\frac{1}{2}\left(-\frac{\mathrm{c}}{\mathrm{a}}\right)\left(-\frac{\mathrm{c}}{\mathrm{b}}\right)=\frac{1}{2} \frac{\mathrm{c}^{2}}{\mathrm{ab}}=\frac{1}{2}\) unit \(\left[\because a, c, b\right.\) are in G.P. \(\left.\Rightarrow c^{2}=a b\right]\)
BITSAT-2020
Co-Ordinate system
88264
The points \(A(-a,-b), B(0,0), C(a, b)\) and \(\mathbf{D}\left(\mathbf{a}^{2}, \mathbf{a b}\right)\) are
1 Vertices of a rectangle
2 Vertices of a parallelogram
3 Vertices of a square
4 Collinear
Explanation:
(D) : Distance between the points \(A(-a,-b)\) and \(\mathrm{B}(0,0)\) \(=\sqrt{(0+a)^{2}+(0+b)^{2}}=\sqrt{a^{2}+b^{2}}\) Distance between the points \(B(0,0)\) and \(\mathrm{C}(\mathrm{a}, \mathrm{b})\) \(\sqrt{(a-0)^{2}+(b-0)^{2}}=\sqrt{a^{2}+b^{2}}\) Distance between the points \(C(a, b)\) and \(D\left(a^{2}, a b\right)\) \(=\sqrt{\left(a^{2}-a\right)^{2}+(a b-b)^{2}}=\sqrt{[a(a-1)]^{2}+[b(a-1)]^{2}}\) \(=\sqrt{a^{2}(a-1)^{2}+b^{2}(a-1)^{2}}=\sqrt{\left(a^{2}+b^{2}\right)(a-1)^{2}}\) \(=(a-1) \sqrt{a^{2}+b^{2}}\) Similarly, distance between the points \(A(-a,-b)\) and \(D\left(a^{2}, a b\right)=\sqrt{\left(a^{2}+a\right)^{2}+(a b+b)^{2}}=(a+1) \sqrt{a^{2}+b^{2}}\) \(A B+B C+C D=\sqrt{a^{2}+b^{2}}+\sqrt{a^{2}+b^{2}}+(a-1) \sqrt{a^{2}+b^{2}}\) \(=(a+1) \sqrt{a^{2}+b^{2}}=A D\) Hence, the point are collinear.
MHT CET-2020
Co-Ordinate system
88265
If \(P(2,2), Q(-2,4)\) and \(R(3,4)\) are the vertices of \(\triangle \mathrm{PQR}\) then the equation of the median through vertex \(R\) is
1 \(x+3 y-9=0\)
2 \(x+3 y+9=0\)
3 \(x-3 y-9=0\)
4 \(x-3 y+9=0\)
Explanation:
(D) : Given, \(P=(2,2) Q=(-2,4) R=(3,4)\) Let \(\mathrm{S}\) be the midpoint of \(\mathrm{PQ}\). \(S=\left(\frac{2+(-2)}{2}, \frac{2+4}{2}\right) \Rightarrow S=(0,3)\) \(\therefore\) Equation of median \(\mathrm{RS}\) is \(\frac{y-3}{3-4}=\frac{x-0}{0-3} \text { i.e. } \frac{y-3}{-1}=\frac{x}{-3}\) \(-3 y+9=-x \text { i.e. } x-3 y+9=0\)
MHT CET-2019
Co-Ordinate system
88266
The minimum area of the triangle formed by the variable line \(3 \cos \theta \cdot x+4 \sin \theta . y=12\) and the co-ordinate axes is
1 144
2 \(25 / 2\)
3 \(49 / 4\)
4 12
Explanation:
(D) : The equation of line formed a triangle with \(\mathrm{X}-\) axis and \(y\)-axis is \(3 \cos \theta \cdot x+4 \sin \theta \cdot y=12 \) \(\frac{x}{(4 / \cos \theta)}+\frac{y}{(3 / \sin \theta)}=1\) Now, The area of triangle AOB will be \(=\frac{1}{2} \times \text { base } \times \text { height }=\frac{1}{2} \times \frac{4}{\cos \theta} \times \frac{3}{\sin \theta}=\frac{12}{\sin 2 \theta}\) The minimum area of the triangle will be when the denominator will be maximum i.e. \(\sin 2 \theta\) will be maximum. Area of \(\triangle \mathrm{AOB}=12\) \(\because \sin 2 \theta\) maximum will be 1
88263
If \(a, c, b\) are in G.P., then the area of the triangle formed by the lines \(a x+b y+c=0\) with the coordinates axes is equal to
1 1
2 2
3 \(\frac{1}{2}\)
4 Nones of these
Explanation:
(C) : Areas of the triangle \(=\frac{1}{2}(\mathrm{x}\) intercept \() \times(\mathrm{y}\) intercept \()\) \(=\frac{1}{2}\left(-\frac{\mathrm{c}}{\mathrm{a}}\right)\left(-\frac{\mathrm{c}}{\mathrm{b}}\right)=\frac{1}{2} \frac{\mathrm{c}^{2}}{\mathrm{ab}}=\frac{1}{2}\) unit \(\left[\because a, c, b\right.\) are in G.P. \(\left.\Rightarrow c^{2}=a b\right]\)
BITSAT-2020
Co-Ordinate system
88264
The points \(A(-a,-b), B(0,0), C(a, b)\) and \(\mathbf{D}\left(\mathbf{a}^{2}, \mathbf{a b}\right)\) are
1 Vertices of a rectangle
2 Vertices of a parallelogram
3 Vertices of a square
4 Collinear
Explanation:
(D) : Distance between the points \(A(-a,-b)\) and \(\mathrm{B}(0,0)\) \(=\sqrt{(0+a)^{2}+(0+b)^{2}}=\sqrt{a^{2}+b^{2}}\) Distance between the points \(B(0,0)\) and \(\mathrm{C}(\mathrm{a}, \mathrm{b})\) \(\sqrt{(a-0)^{2}+(b-0)^{2}}=\sqrt{a^{2}+b^{2}}\) Distance between the points \(C(a, b)\) and \(D\left(a^{2}, a b\right)\) \(=\sqrt{\left(a^{2}-a\right)^{2}+(a b-b)^{2}}=\sqrt{[a(a-1)]^{2}+[b(a-1)]^{2}}\) \(=\sqrt{a^{2}(a-1)^{2}+b^{2}(a-1)^{2}}=\sqrt{\left(a^{2}+b^{2}\right)(a-1)^{2}}\) \(=(a-1) \sqrt{a^{2}+b^{2}}\) Similarly, distance between the points \(A(-a,-b)\) and \(D\left(a^{2}, a b\right)=\sqrt{\left(a^{2}+a\right)^{2}+(a b+b)^{2}}=(a+1) \sqrt{a^{2}+b^{2}}\) \(A B+B C+C D=\sqrt{a^{2}+b^{2}}+\sqrt{a^{2}+b^{2}}+(a-1) \sqrt{a^{2}+b^{2}}\) \(=(a+1) \sqrt{a^{2}+b^{2}}=A D\) Hence, the point are collinear.
MHT CET-2020
Co-Ordinate system
88265
If \(P(2,2), Q(-2,4)\) and \(R(3,4)\) are the vertices of \(\triangle \mathrm{PQR}\) then the equation of the median through vertex \(R\) is
1 \(x+3 y-9=0\)
2 \(x+3 y+9=0\)
3 \(x-3 y-9=0\)
4 \(x-3 y+9=0\)
Explanation:
(D) : Given, \(P=(2,2) Q=(-2,4) R=(3,4)\) Let \(\mathrm{S}\) be the midpoint of \(\mathrm{PQ}\). \(S=\left(\frac{2+(-2)}{2}, \frac{2+4}{2}\right) \Rightarrow S=(0,3)\) \(\therefore\) Equation of median \(\mathrm{RS}\) is \(\frac{y-3}{3-4}=\frac{x-0}{0-3} \text { i.e. } \frac{y-3}{-1}=\frac{x}{-3}\) \(-3 y+9=-x \text { i.e. } x-3 y+9=0\)
MHT CET-2019
Co-Ordinate system
88266
The minimum area of the triangle formed by the variable line \(3 \cos \theta \cdot x+4 \sin \theta . y=12\) and the co-ordinate axes is
1 144
2 \(25 / 2\)
3 \(49 / 4\)
4 12
Explanation:
(D) : The equation of line formed a triangle with \(\mathrm{X}-\) axis and \(y\)-axis is \(3 \cos \theta \cdot x+4 \sin \theta \cdot y=12 \) \(\frac{x}{(4 / \cos \theta)}+\frac{y}{(3 / \sin \theta)}=1\) Now, The area of triangle AOB will be \(=\frac{1}{2} \times \text { base } \times \text { height }=\frac{1}{2} \times \frac{4}{\cos \theta} \times \frac{3}{\sin \theta}=\frac{12}{\sin 2 \theta}\) The minimum area of the triangle will be when the denominator will be maximum i.e. \(\sin 2 \theta\) will be maximum. Area of \(\triangle \mathrm{AOB}=12\) \(\because \sin 2 \theta\) maximum will be 1
88263
If \(a, c, b\) are in G.P., then the area of the triangle formed by the lines \(a x+b y+c=0\) with the coordinates axes is equal to
1 1
2 2
3 \(\frac{1}{2}\)
4 Nones of these
Explanation:
(C) : Areas of the triangle \(=\frac{1}{2}(\mathrm{x}\) intercept \() \times(\mathrm{y}\) intercept \()\) \(=\frac{1}{2}\left(-\frac{\mathrm{c}}{\mathrm{a}}\right)\left(-\frac{\mathrm{c}}{\mathrm{b}}\right)=\frac{1}{2} \frac{\mathrm{c}^{2}}{\mathrm{ab}}=\frac{1}{2}\) unit \(\left[\because a, c, b\right.\) are in G.P. \(\left.\Rightarrow c^{2}=a b\right]\)
BITSAT-2020
Co-Ordinate system
88264
The points \(A(-a,-b), B(0,0), C(a, b)\) and \(\mathbf{D}\left(\mathbf{a}^{2}, \mathbf{a b}\right)\) are
1 Vertices of a rectangle
2 Vertices of a parallelogram
3 Vertices of a square
4 Collinear
Explanation:
(D) : Distance between the points \(A(-a,-b)\) and \(\mathrm{B}(0,0)\) \(=\sqrt{(0+a)^{2}+(0+b)^{2}}=\sqrt{a^{2}+b^{2}}\) Distance between the points \(B(0,0)\) and \(\mathrm{C}(\mathrm{a}, \mathrm{b})\) \(\sqrt{(a-0)^{2}+(b-0)^{2}}=\sqrt{a^{2}+b^{2}}\) Distance between the points \(C(a, b)\) and \(D\left(a^{2}, a b\right)\) \(=\sqrt{\left(a^{2}-a\right)^{2}+(a b-b)^{2}}=\sqrt{[a(a-1)]^{2}+[b(a-1)]^{2}}\) \(=\sqrt{a^{2}(a-1)^{2}+b^{2}(a-1)^{2}}=\sqrt{\left(a^{2}+b^{2}\right)(a-1)^{2}}\) \(=(a-1) \sqrt{a^{2}+b^{2}}\) Similarly, distance between the points \(A(-a,-b)\) and \(D\left(a^{2}, a b\right)=\sqrt{\left(a^{2}+a\right)^{2}+(a b+b)^{2}}=(a+1) \sqrt{a^{2}+b^{2}}\) \(A B+B C+C D=\sqrt{a^{2}+b^{2}}+\sqrt{a^{2}+b^{2}}+(a-1) \sqrt{a^{2}+b^{2}}\) \(=(a+1) \sqrt{a^{2}+b^{2}}=A D\) Hence, the point are collinear.
MHT CET-2020
Co-Ordinate system
88265
If \(P(2,2), Q(-2,4)\) and \(R(3,4)\) are the vertices of \(\triangle \mathrm{PQR}\) then the equation of the median through vertex \(R\) is
1 \(x+3 y-9=0\)
2 \(x+3 y+9=0\)
3 \(x-3 y-9=0\)
4 \(x-3 y+9=0\)
Explanation:
(D) : Given, \(P=(2,2) Q=(-2,4) R=(3,4)\) Let \(\mathrm{S}\) be the midpoint of \(\mathrm{PQ}\). \(S=\left(\frac{2+(-2)}{2}, \frac{2+4}{2}\right) \Rightarrow S=(0,3)\) \(\therefore\) Equation of median \(\mathrm{RS}\) is \(\frac{y-3}{3-4}=\frac{x-0}{0-3} \text { i.e. } \frac{y-3}{-1}=\frac{x}{-3}\) \(-3 y+9=-x \text { i.e. } x-3 y+9=0\)
MHT CET-2019
Co-Ordinate system
88266
The minimum area of the triangle formed by the variable line \(3 \cos \theta \cdot x+4 \sin \theta . y=12\) and the co-ordinate axes is
1 144
2 \(25 / 2\)
3 \(49 / 4\)
4 12
Explanation:
(D) : The equation of line formed a triangle with \(\mathrm{X}-\) axis and \(y\)-axis is \(3 \cos \theta \cdot x+4 \sin \theta \cdot y=12 \) \(\frac{x}{(4 / \cos \theta)}+\frac{y}{(3 / \sin \theta)}=1\) Now, The area of triangle AOB will be \(=\frac{1}{2} \times \text { base } \times \text { height }=\frac{1}{2} \times \frac{4}{\cos \theta} \times \frac{3}{\sin \theta}=\frac{12}{\sin 2 \theta}\) The minimum area of the triangle will be when the denominator will be maximum i.e. \(\sin 2 \theta\) will be maximum. Area of \(\triangle \mathrm{AOB}=12\) \(\because \sin 2 \theta\) maximum will be 1
88263
If \(a, c, b\) are in G.P., then the area of the triangle formed by the lines \(a x+b y+c=0\) with the coordinates axes is equal to
1 1
2 2
3 \(\frac{1}{2}\)
4 Nones of these
Explanation:
(C) : Areas of the triangle \(=\frac{1}{2}(\mathrm{x}\) intercept \() \times(\mathrm{y}\) intercept \()\) \(=\frac{1}{2}\left(-\frac{\mathrm{c}}{\mathrm{a}}\right)\left(-\frac{\mathrm{c}}{\mathrm{b}}\right)=\frac{1}{2} \frac{\mathrm{c}^{2}}{\mathrm{ab}}=\frac{1}{2}\) unit \(\left[\because a, c, b\right.\) are in G.P. \(\left.\Rightarrow c^{2}=a b\right]\)
BITSAT-2020
Co-Ordinate system
88264
The points \(A(-a,-b), B(0,0), C(a, b)\) and \(\mathbf{D}\left(\mathbf{a}^{2}, \mathbf{a b}\right)\) are
1 Vertices of a rectangle
2 Vertices of a parallelogram
3 Vertices of a square
4 Collinear
Explanation:
(D) : Distance between the points \(A(-a,-b)\) and \(\mathrm{B}(0,0)\) \(=\sqrt{(0+a)^{2}+(0+b)^{2}}=\sqrt{a^{2}+b^{2}}\) Distance between the points \(B(0,0)\) and \(\mathrm{C}(\mathrm{a}, \mathrm{b})\) \(\sqrt{(a-0)^{2}+(b-0)^{2}}=\sqrt{a^{2}+b^{2}}\) Distance between the points \(C(a, b)\) and \(D\left(a^{2}, a b\right)\) \(=\sqrt{\left(a^{2}-a\right)^{2}+(a b-b)^{2}}=\sqrt{[a(a-1)]^{2}+[b(a-1)]^{2}}\) \(=\sqrt{a^{2}(a-1)^{2}+b^{2}(a-1)^{2}}=\sqrt{\left(a^{2}+b^{2}\right)(a-1)^{2}}\) \(=(a-1) \sqrt{a^{2}+b^{2}}\) Similarly, distance between the points \(A(-a,-b)\) and \(D\left(a^{2}, a b\right)=\sqrt{\left(a^{2}+a\right)^{2}+(a b+b)^{2}}=(a+1) \sqrt{a^{2}+b^{2}}\) \(A B+B C+C D=\sqrt{a^{2}+b^{2}}+\sqrt{a^{2}+b^{2}}+(a-1) \sqrt{a^{2}+b^{2}}\) \(=(a+1) \sqrt{a^{2}+b^{2}}=A D\) Hence, the point are collinear.
MHT CET-2020
Co-Ordinate system
88265
If \(P(2,2), Q(-2,4)\) and \(R(3,4)\) are the vertices of \(\triangle \mathrm{PQR}\) then the equation of the median through vertex \(R\) is
1 \(x+3 y-9=0\)
2 \(x+3 y+9=0\)
3 \(x-3 y-9=0\)
4 \(x-3 y+9=0\)
Explanation:
(D) : Given, \(P=(2,2) Q=(-2,4) R=(3,4)\) Let \(\mathrm{S}\) be the midpoint of \(\mathrm{PQ}\). \(S=\left(\frac{2+(-2)}{2}, \frac{2+4}{2}\right) \Rightarrow S=(0,3)\) \(\therefore\) Equation of median \(\mathrm{RS}\) is \(\frac{y-3}{3-4}=\frac{x-0}{0-3} \text { i.e. } \frac{y-3}{-1}=\frac{x}{-3}\) \(-3 y+9=-x \text { i.e. } x-3 y+9=0\)
MHT CET-2019
Co-Ordinate system
88266
The minimum area of the triangle formed by the variable line \(3 \cos \theta \cdot x+4 \sin \theta . y=12\) and the co-ordinate axes is
1 144
2 \(25 / 2\)
3 \(49 / 4\)
4 12
Explanation:
(D) : The equation of line formed a triangle with \(\mathrm{X}-\) axis and \(y\)-axis is \(3 \cos \theta \cdot x+4 \sin \theta \cdot y=12 \) \(\frac{x}{(4 / \cos \theta)}+\frac{y}{(3 / \sin \theta)}=1\) Now, The area of triangle AOB will be \(=\frac{1}{2} \times \text { base } \times \text { height }=\frac{1}{2} \times \frac{4}{\cos \theta} \times \frac{3}{\sin \theta}=\frac{12}{\sin 2 \theta}\) The minimum area of the triangle will be when the denominator will be maximum i.e. \(\sin 2 \theta\) will be maximum. Area of \(\triangle \mathrm{AOB}=12\) \(\because \sin 2 \theta\) maximum will be 1
