QBManager

Co-Ordinate system

88206 If the straight lines $2 x + 3 y - 3 = 0$ and $x + k y +$ $7 = 0$ are perpendicular, then the value of $k$ is

1

2 / 3

2

3 / 2

3

- 2 / 3

4

- 3 / 2

Explanation:

(C) : Given,
Equation of straight lines are
$\begin{aligned} (i) & 2 x + 3 y - 3 = 0 \\ (ii) & x + k y + 7 = 0 \end{aligned}$
Equation (i) and (ii) can be written as shown below
$y = - \frac{2}{3} x + 1$
$∴$ Slope $(m_{1}) = - \frac{2}{3}$
And, $y = - \frac{1}{k} x - \frac{7}{k}$
Slope $(m_{2}) = - \frac{1}{k}$
We know that both the straight line are perpendicular to each other.
$∴ m_{1} m_{2} = - 1$
$- \frac{2}{3} \times (- \frac{1}{k}) = - 1, k = - \frac{2}{3}$

Karnataka CET-2016

Co-Ordinate system

88207 A straight line passes through the points $(5, 0)$ and $(0, 3)$ . The length of perpendicular from the point $(4, 4)$ on the line is

1

\frac{15}{\sqrt{34}}

2

\frac{\sqrt{17}}{2}

3

\frac{17}{2}

4

\sqrt{\frac{17}{2}}

Explanation:

(D) : We know,
The equation Straight line passes through point $(5, 0)$ and $(0, 3)$
$(y - y_{1}) = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} (x - x_{1})$
$(y - 0) = \frac{3 - 0}{0 - 5} (x - 5)$
$y = - \frac{3}{5} (x - 5)$
$5 y = - 3 x + 15$
$3 x + 5 y - 15 = 0$
Therefore, the distance of this line from the point $(4, 4)$ is
$d = \frac{| 3 \times 4 + 5 \times 4 - 15 |}{\sqrt{3^{2} + 5^{2}}} = \frac{17}{\sqrt{34}}$
$d = \frac{\sqrt{17} \times \sqrt{17}}{\sqrt{17} \times \sqrt{2}} = \sqrt{\frac{17}{2}}$

Karnataka CET-2014

Co-Ordinate system

88208 Find the transformed equation of the straight line $x y - x - y + 1 = 0$ , when the origin is shifted to the point $(1, 1)$ after translation of axis.

1

x y = 5

2

x y = 2

3

x y = 0

4

x y = 8

Explanation:

(C) : Let the coordinates of a point $p$ changes from $(x, y)$ to $(x^{'}, y^{'})$ in new coordinates axis where origin has the coordinates $h = 1, k = 1$
Then, $x = x^{'} + 1, y = y^{'} + 1$ .
Substituting these values in the given equation of straight line
$(x^{'} + 1) (y^{'} + 1) - (x^{'} + 1) - (y^{'} + 1) + 1 = 0$
$x^{'} y^{'} + x^{'} + y^{'} + 1 - x^{'} - 1 - y^{'} - 1 + 1 = 0$
$x^{'} y^{'} = 0$
Therefore, the equation of straight line in the new system is $xy = 0$ .

COMEDK-2017

Co-Ordinate system

88209 If the axis are shifted to the point $(1, - 2)$ without solution, then the equation $2 x^{2} + y^{2} - 4 x + 4 y = 0$ becomes

1

2 X^{2} + 3 Y^{2} = 6

2

2 X^{2} + Y^{2} = 6

3

X^{2} + 2 Y^{2} = 6

4 None of these

Explanation:

(B) :Given the point $(1, - 2)$
The equation is $2 x^{2} + y^{2} - 4 x + 4 y = 0$
Now, substituting $x = X + 1$ and $y = Y - 2$ in given equation, we get
$2 (X + 1)^{2} + (Y - 2)^{2} - 4 (X + 1) + 4 (Y - 2) = 0$
$2 X^{2} + Y^{2} = 6$

VITEEE-2011

Co-Ordinate system

88206 If the straight lines $2 x + 3 y - 3 = 0$ and $x + k y +$ $7 = 0$ are perpendicular, then the value of $k$ is

1

2 / 3

2

3 / 2

3

- 2 / 3

4

- 3 / 2

Explanation:

(C) : Given,
Equation of straight lines are
$\begin{aligned} (i) & 2 x + 3 y - 3 = 0 \\ (ii) & x + k y + 7 = 0 \end{aligned}$
Equation (i) and (ii) can be written as shown below
$y = - \frac{2}{3} x + 1$
$∴$ Slope $(m_{1}) = - \frac{2}{3}$
And, $y = - \frac{1}{k} x - \frac{7}{k}$
Slope $(m_{2}) = - \frac{1}{k}$
We know that both the straight line are perpendicular to each other.
$∴ m_{1} m_{2} = - 1$
$- \frac{2}{3} \times (- \frac{1}{k}) = - 1, k = - \frac{2}{3}$

Karnataka CET-2016

Co-Ordinate system

88207 A straight line passes through the points $(5, 0)$ and $(0, 3)$ . The length of perpendicular from the point $(4, 4)$ on the line is

1

\frac{15}{\sqrt{34}}

2

\frac{\sqrt{17}}{2}

3

\frac{17}{2}

4

\sqrt{\frac{17}{2}}

Explanation:

(D) : We know,
The equation Straight line passes through point $(5, 0)$ and $(0, 3)$
$(y - y_{1}) = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} (x - x_{1})$
$(y - 0) = \frac{3 - 0}{0 - 5} (x - 5)$
$y = - \frac{3}{5} (x - 5)$
$5 y = - 3 x + 15$
$3 x + 5 y - 15 = 0$
Therefore, the distance of this line from the point $(4, 4)$ is
$d = \frac{| 3 \times 4 + 5 \times 4 - 15 |}{\sqrt{3^{2} + 5^{2}}} = \frac{17}{\sqrt{34}}$
$d = \frac{\sqrt{17} \times \sqrt{17}}{\sqrt{17} \times \sqrt{2}} = \sqrt{\frac{17}{2}}$

Karnataka CET-2014

Co-Ordinate system

88208 Find the transformed equation of the straight line $x y - x - y + 1 = 0$ , when the origin is shifted to the point $(1, 1)$ after translation of axis.

1

x y = 5

2

x y = 2

3

x y = 0

4

x y = 8

Explanation:

(C) : Let the coordinates of a point $p$ changes from $(x, y)$ to $(x^{'}, y^{'})$ in new coordinates axis where origin has the coordinates $h = 1, k = 1$
Then, $x = x^{'} + 1, y = y^{'} + 1$ .
Substituting these values in the given equation of straight line
$(x^{'} + 1) (y^{'} + 1) - (x^{'} + 1) - (y^{'} + 1) + 1 = 0$
$x^{'} y^{'} + x^{'} + y^{'} + 1 - x^{'} - 1 - y^{'} - 1 + 1 = 0$
$x^{'} y^{'} = 0$
Therefore, the equation of straight line in the new system is $xy = 0$ .

COMEDK-2017

Co-Ordinate system

88209 If the axis are shifted to the point $(1, - 2)$ without solution, then the equation $2 x^{2} + y^{2} - 4 x + 4 y = 0$ becomes

1

2 X^{2} + 3 Y^{2} = 6

2

2 X^{2} + Y^{2} = 6

3

X^{2} + 2 Y^{2} = 6

4 None of these

Explanation:

(B) :Given the point $(1, - 2)$
The equation is $2 x^{2} + y^{2} - 4 x + 4 y = 0$
Now, substituting $x = X + 1$ and $y = Y - 2$ in given equation, we get
$2 (X + 1)^{2} + (Y - 2)^{2} - 4 (X + 1) + 4 (Y - 2) = 0$
$2 X^{2} + Y^{2} = 6$

VITEEE-2011

Co-Ordinate system

88206 If the straight lines $2 x + 3 y - 3 = 0$ and $x + k y +$ $7 = 0$ are perpendicular, then the value of $k$ is

1

2 / 3

2

3 / 2

3

- 2 / 3

4

- 3 / 2

Explanation:

(C) : Given,
Equation of straight lines are
$\begin{aligned} (i) & 2 x + 3 y - 3 = 0 \\ (ii) & x + k y + 7 = 0 \end{aligned}$
Equation (i) and (ii) can be written as shown below
$y = - \frac{2}{3} x + 1$
$∴$ Slope $(m_{1}) = - \frac{2}{3}$
And, $y = - \frac{1}{k} x - \frac{7}{k}$
Slope $(m_{2}) = - \frac{1}{k}$
We know that both the straight line are perpendicular to each other.
$∴ m_{1} m_{2} = - 1$
$- \frac{2}{3} \times (- \frac{1}{k}) = - 1, k = - \frac{2}{3}$

Karnataka CET-2016

Co-Ordinate system

88207 A straight line passes through the points $(5, 0)$ and $(0, 3)$ . The length of perpendicular from the point $(4, 4)$ on the line is

1

\frac{15}{\sqrt{34}}

2

\frac{\sqrt{17}}{2}

3

\frac{17}{2}

4

\sqrt{\frac{17}{2}}

Explanation:

(D) : We know,
The equation Straight line passes through point $(5, 0)$ and $(0, 3)$
$(y - y_{1}) = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} (x - x_{1})$
$(y - 0) = \frac{3 - 0}{0 - 5} (x - 5)$
$y = - \frac{3}{5} (x - 5)$
$5 y = - 3 x + 15$
$3 x + 5 y - 15 = 0$
Therefore, the distance of this line from the point $(4, 4)$ is
$d = \frac{| 3 \times 4 + 5 \times 4 - 15 |}{\sqrt{3^{2} + 5^{2}}} = \frac{17}{\sqrt{34}}$
$d = \frac{\sqrt{17} \times \sqrt{17}}{\sqrt{17} \times \sqrt{2}} = \sqrt{\frac{17}{2}}$

Karnataka CET-2014

Co-Ordinate system

88208 Find the transformed equation of the straight line $x y - x - y + 1 = 0$ , when the origin is shifted to the point $(1, 1)$ after translation of axis.

1

x y = 5

2

x y = 2

3

x y = 0

4

x y = 8

Explanation:

(C) : Let the coordinates of a point $p$ changes from $(x, y)$ to $(x^{'}, y^{'})$ in new coordinates axis where origin has the coordinates $h = 1, k = 1$
Then, $x = x^{'} + 1, y = y^{'} + 1$ .
Substituting these values in the given equation of straight line
$(x^{'} + 1) (y^{'} + 1) - (x^{'} + 1) - (y^{'} + 1) + 1 = 0$
$x^{'} y^{'} + x^{'} + y^{'} + 1 - x^{'} - 1 - y^{'} - 1 + 1 = 0$
$x^{'} y^{'} = 0$
Therefore, the equation of straight line in the new system is $xy = 0$ .

COMEDK-2017

Co-Ordinate system

88209 If the axis are shifted to the point $(1, - 2)$ without solution, then the equation $2 x^{2} + y^{2} - 4 x + 4 y = 0$ becomes

1

2 X^{2} + 3 Y^{2} = 6

2

2 X^{2} + Y^{2} = 6

3

X^{2} + 2 Y^{2} = 6

4 None of these

Explanation:

(B) :Given the point $(1, - 2)$
The equation is $2 x^{2} + y^{2} - 4 x + 4 y = 0$
Now, substituting $x = X + 1$ and $y = Y - 2$ in given equation, we get
$2 (X + 1)^{2} + (Y - 2)^{2} - 4 (X + 1) + 4 (Y - 2) = 0$
$2 X^{2} + Y^{2} = 6$

VITEEE-2011

Co-Ordinate system

88206 If the straight lines $2 x + 3 y - 3 = 0$ and $x + k y +$ $7 = 0$ are perpendicular, then the value of $k$ is

1

2 / 3

2

3 / 2

3

- 2 / 3

4

- 3 / 2

Explanation:

(C) : Given,
Equation of straight lines are
$\begin{aligned} (i) & 2 x + 3 y - 3 = 0 \\ (ii) & x + k y + 7 = 0 \end{aligned}$
Equation (i) and (ii) can be written as shown below
$y = - \frac{2}{3} x + 1$
$∴$ Slope $(m_{1}) = - \frac{2}{3}$
And, $y = - \frac{1}{k} x - \frac{7}{k}$
Slope $(m_{2}) = - \frac{1}{k}$
We know that both the straight line are perpendicular to each other.
$∴ m_{1} m_{2} = - 1$
$- \frac{2}{3} \times (- \frac{1}{k}) = - 1, k = - \frac{2}{3}$

Karnataka CET-2016

Co-Ordinate system

88207 A straight line passes through the points $(5, 0)$ and $(0, 3)$ . The length of perpendicular from the point $(4, 4)$ on the line is

1

\frac{15}{\sqrt{34}}

2

\frac{\sqrt{17}}{2}

3

\frac{17}{2}

4

\sqrt{\frac{17}{2}}

Explanation:

(D) : We know,
The equation Straight line passes through point $(5, 0)$ and $(0, 3)$
$(y - y_{1}) = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} (x - x_{1})$
$(y - 0) = \frac{3 - 0}{0 - 5} (x - 5)$
$y = - \frac{3}{5} (x - 5)$
$5 y = - 3 x + 15$
$3 x + 5 y - 15 = 0$
Therefore, the distance of this line from the point $(4, 4)$ is
$d = \frac{| 3 \times 4 + 5 \times 4 - 15 |}{\sqrt{3^{2} + 5^{2}}} = \frac{17}{\sqrt{34}}$
$d = \frac{\sqrt{17} \times \sqrt{17}}{\sqrt{17} \times \sqrt{2}} = \sqrt{\frac{17}{2}}$

Karnataka CET-2014

Co-Ordinate system

88208 Find the transformed equation of the straight line $x y - x - y + 1 = 0$ , when the origin is shifted to the point $(1, 1)$ after translation of axis.

1

x y = 5

2

x y = 2

3

x y = 0

4

x y = 8

Explanation:

(C) : Let the coordinates of a point $p$ changes from $(x, y)$ to $(x^{'}, y^{'})$ in new coordinates axis where origin has the coordinates $h = 1, k = 1$
Then, $x = x^{'} + 1, y = y^{'} + 1$ .
Substituting these values in the given equation of straight line
$(x^{'} + 1) (y^{'} + 1) - (x^{'} + 1) - (y^{'} + 1) + 1 = 0$
$x^{'} y^{'} + x^{'} + y^{'} + 1 - x^{'} - 1 - y^{'} - 1 + 1 = 0$
$x^{'} y^{'} = 0$
Therefore, the equation of straight line in the new system is $xy = 0$ .

COMEDK-2017

Co-Ordinate system

88209 If the axis are shifted to the point $(1, - 2)$ without solution, then the equation $2 x^{2} + y^{2} - 4 x + 4 y = 0$ becomes

1

2 X^{2} + 3 Y^{2} = 6

2

2 X^{2} + Y^{2} = 6

3

X^{2} + 2 Y^{2} = 6

4 None of these

Explanation:

(B) :Given the point $(1, - 2)$
The equation is $2 x^{2} + y^{2} - 4 x + 4 y = 0$
Now, substituting $x = X + 1$ and $y = Y - 2$ in given equation, we get
$2 (X + 1)^{2} + (Y - 2)^{2} - 4 (X + 1) + 4 (Y - 2) = 0$
$2 X^{2} + Y^{2} = 6$

VITEEE-2011