QBManager

Co-Ordinate system

88283 The coordinates of the point on the line through the points $A (3, 4, 1)$ and $B (5, 1, 6)$ crosses XY-plane are

1

(\frac{13}{5}, \frac{23}{5}, 0)

2

(\frac{3}{5}, \frac{2}{5}, 0)

3

(1, 1, 0)

4

(- \frac{13}{5}, \frac{23}{5}, 0)

Explanation:

(A) : Given,
The co-ordinate of the points $A (3, 4, 1)$ and $B (5, 1, 6)$
Now,
The equation of line in space can be written as
$\frac{x - x_{1}}{x_{2} - x_{1}} = \frac{y - y_{1}}{y_{2} - y_{1}} = \frac{z - z_{1}}{z_{2} - z_{1}} [\begin{array}{l} Where (x_{1}, y_{1}, z_{1}) and \\ (x_{2}, y_{2}, z_{2}) are the points \end{array}]$
$\frac{x - 3}{5 - 3} = \frac{y - 4}{1 - 4} = \frac{z - 1}{6 - 1}$
$\frac{x - 3}{2} = \frac{y - 4}{- 3} = \frac{z - 1}{5} = k$
Now, In xy plane $z = 0$
$k = - \frac{1}{5}$
$∴ \frac{x - 3}{2} = \frac{- 1}{5} and \frac{y - 4}{- 3} = - \frac{1}{5}$
$x - 3 = - \frac{2}{5} and y - 4 = 3 / 5$
$x = - \frac{2}{5} + 3, y = \frac{3}{5} + 4$
$x = \frac{13}{5} and y = \frac{23}{5}$
Hence, the required co-ordinate $(\frac{13}{5}, \frac{23}{5}, 0)$

[JCECE-2018]

Co-Ordinate system

88284 If $y = f (x)$ makes positive intercepts of 2 and 1 unit an $x$ and $y$ coordinate axes and encloses an area of $\frac{3}{4}$ square unit with the axes, then $\int_{0}^{2} x f^{'} (x) d x$ is is

1

\frac{3}{2}

2 1

3

\frac{5}{4}

4

- \frac{3}{4}

Explanation:

(D) : Given,
$\int_{0}^{2} f (x) d x = 3 / 4$
$∴ \int_{0}^{2} x f^{'} (x) d x$
$= x \int_{0}^{2} f^{'} (x) d x - \int_{0}^{2} f (x) d x$

$= {[x \int_{0}^{2} f (x) dx]}_{0}^{2} - 3 / 4 = 2 f (2) - 3 / 4$
$= 2 \times 0 - 3 / 4$
$[y = f (x) \Rightarrow y = f (2)] = - 3 / 4$

BCECE-2018

Co-Ordinate system

88285 If $A (2, 3)$ and $B (- 2, 1)$ are two vertices of a triangle and third vertex moves on the line $2 x + 3 y = 9$ , then the locus of the centroid of the triangle is

1

2 x + 3 y = 1

2

2 x + y = 3

3

2 x - 3 y = 1

4

x - y = 1

Explanation:

(A) : Given,
The co-ordinate of vertices of triangle are $A (2, - 3), B (-$ $2, 1)$ and $C (x, y)$
Now, Let the co-ordinate of centroid be $(h, k)$
$∴ h = (\frac{2 - 2 + x}{3}) \Rightarrow x = 3 h$
$k = (\frac{- 3 + 1 + y}{3}) \Rightarrow y = 3 k + 2$
Where $(x, y)$ are the point on the line $2 x + 3 y = 9$
$∴ 2 (3 h) + 3 (3 k + 2) = 9$
$6 h + 9 k + 6 = 9$
$6 h + 9 k = 3$
$2 h + 3 k = 1$
Locus of the centroid of the triangle is $2 x + 3 y = 1$

BCECE-2017

Co-Ordinate system

88287 Let $A (2, - 3)$ and $B (- 2, 1)$ be vertices of a $△ ABC$ . If The centroid of this triangle moves on the line $2 x + 3 y = 1$ , then the locus of the vertex $C$ is the line

1

2 x + 3 y = 9

2

2 x - 3 y = 7

3

3 x + 2 y = 5

4

3 x - 2 y = 3

Explanation:

(A) : Given,
The co-ordinate of vertices of triangle $A (2, - 3), B (- 2, 1)$ and $C (h, k)$
Now, co-ordinate of centroid lies on the line $2 x + 3 y = 1$

$\Rightarrow (\frac{2 - 2 + h}{3}, \frac{- 3 + 1 + k}{3}) \Rightarrow (\frac{h}{3}, \frac{- 2 + k}{3})$
$∴ (\frac{h}{3}, \frac{- 2 + k}{3})$ satisfy the equation ${2 x + 3 y = 1}$
$\Rightarrow 2 (\frac{h}{3}) + 3 (\frac{- 2 + k}{3}) = 1$
$\frac{2}{3} h + (- 2 + k) = 1$
$\Rightarrow 2 h - 6 + 3 k = 3$
$2 h + 3 k = 9$
Locus of the vertex $C$ is the line $2 x + 3 y = 9$

BCECE-2014

Co-Ordinate system

88283 The coordinates of the point on the line through the points $A (3, 4, 1)$ and $B (5, 1, 6)$ crosses XY-plane are

1

(\frac{13}{5}, \frac{23}{5}, 0)

2

(\frac{3}{5}, \frac{2}{5}, 0)

3

(1, 1, 0)

4

(- \frac{13}{5}, \frac{23}{5}, 0)

Explanation:

(A) : Given,
The co-ordinate of the points $A (3, 4, 1)$ and $B (5, 1, 6)$
Now,
The equation of line in space can be written as
$\frac{x - x_{1}}{x_{2} - x_{1}} = \frac{y - y_{1}}{y_{2} - y_{1}} = \frac{z - z_{1}}{z_{2} - z_{1}} [\begin{array}{l} Where (x_{1}, y_{1}, z_{1}) and \\ (x_{2}, y_{2}, z_{2}) are the points \end{array}]$
$\frac{x - 3}{5 - 3} = \frac{y - 4}{1 - 4} = \frac{z - 1}{6 - 1}$
$\frac{x - 3}{2} = \frac{y - 4}{- 3} = \frac{z - 1}{5} = k$
Now, In xy plane $z = 0$
$k = - \frac{1}{5}$
$∴ \frac{x - 3}{2} = \frac{- 1}{5} and \frac{y - 4}{- 3} = - \frac{1}{5}$
$x - 3 = - \frac{2}{5} and y - 4 = 3 / 5$
$x = - \frac{2}{5} + 3, y = \frac{3}{5} + 4$
$x = \frac{13}{5} and y = \frac{23}{5}$
Hence, the required co-ordinate $(\frac{13}{5}, \frac{23}{5}, 0)$

[JCECE-2018]

Co-Ordinate system

88284 If $y = f (x)$ makes positive intercepts of 2 and 1 unit an $x$ and $y$ coordinate axes and encloses an area of $\frac{3}{4}$ square unit with the axes, then $\int_{0}^{2} x f^{'} (x) d x$ is is

1

\frac{3}{2}

2 1

3

\frac{5}{4}

4

- \frac{3}{4}

Explanation:

(D) : Given,
$\int_{0}^{2} f (x) d x = 3 / 4$
$∴ \int_{0}^{2} x f^{'} (x) d x$
$= x \int_{0}^{2} f^{'} (x) d x - \int_{0}^{2} f (x) d x$

$= {[x \int_{0}^{2} f (x) dx]}_{0}^{2} - 3 / 4 = 2 f (2) - 3 / 4$
$= 2 \times 0 - 3 / 4$
$[y = f (x) \Rightarrow y = f (2)] = - 3 / 4$

BCECE-2018

Co-Ordinate system

88285 If $A (2, 3)$ and $B (- 2, 1)$ are two vertices of a triangle and third vertex moves on the line $2 x + 3 y = 9$ , then the locus of the centroid of the triangle is

1

2 x + 3 y = 1

2

2 x + y = 3

3

2 x - 3 y = 1

4

x - y = 1

Explanation:

(A) : Given,
The co-ordinate of vertices of triangle are $A (2, - 3), B (-$ $2, 1)$ and $C (x, y)$
Now, Let the co-ordinate of centroid be $(h, k)$
$∴ h = (\frac{2 - 2 + x}{3}) \Rightarrow x = 3 h$
$k = (\frac{- 3 + 1 + y}{3}) \Rightarrow y = 3 k + 2$
Where $(x, y)$ are the point on the line $2 x + 3 y = 9$
$∴ 2 (3 h) + 3 (3 k + 2) = 9$
$6 h + 9 k + 6 = 9$
$6 h + 9 k = 3$
$2 h + 3 k = 1$
Locus of the centroid of the triangle is $2 x + 3 y = 1$

BCECE-2017

Co-Ordinate system

88287 Let $A (2, - 3)$ and $B (- 2, 1)$ be vertices of a $△ ABC$ . If The centroid of this triangle moves on the line $2 x + 3 y = 1$ , then the locus of the vertex $C$ is the line

1

2 x + 3 y = 9

2

2 x - 3 y = 7

3

3 x + 2 y = 5

4

3 x - 2 y = 3

Explanation:

(A) : Given,
The co-ordinate of vertices of triangle $A (2, - 3), B (- 2, 1)$ and $C (h, k)$
Now, co-ordinate of centroid lies on the line $2 x + 3 y = 1$

$\Rightarrow (\frac{2 - 2 + h}{3}, \frac{- 3 + 1 + k}{3}) \Rightarrow (\frac{h}{3}, \frac{- 2 + k}{3})$
$∴ (\frac{h}{3}, \frac{- 2 + k}{3})$ satisfy the equation ${2 x + 3 y = 1}$
$\Rightarrow 2 (\frac{h}{3}) + 3 (\frac{- 2 + k}{3}) = 1$
$\frac{2}{3} h + (- 2 + k) = 1$
$\Rightarrow 2 h - 6 + 3 k = 3$
$2 h + 3 k = 9$
Locus of the vertex $C$ is the line $2 x + 3 y = 9$

BCECE-2014

Co-Ordinate system

88283 The coordinates of the point on the line through the points $A (3, 4, 1)$ and $B (5, 1, 6)$ crosses XY-plane are

1

(\frac{13}{5}, \frac{23}{5}, 0)

2

(\frac{3}{5}, \frac{2}{5}, 0)

3

(1, 1, 0)

4

(- \frac{13}{5}, \frac{23}{5}, 0)

Explanation:

(A) : Given,
The co-ordinate of the points $A (3, 4, 1)$ and $B (5, 1, 6)$
Now,
The equation of line in space can be written as
$\frac{x - x_{1}}{x_{2} - x_{1}} = \frac{y - y_{1}}{y_{2} - y_{1}} = \frac{z - z_{1}}{z_{2} - z_{1}} [\begin{array}{l} Where (x_{1}, y_{1}, z_{1}) and \\ (x_{2}, y_{2}, z_{2}) are the points \end{array}]$
$\frac{x - 3}{5 - 3} = \frac{y - 4}{1 - 4} = \frac{z - 1}{6 - 1}$
$\frac{x - 3}{2} = \frac{y - 4}{- 3} = \frac{z - 1}{5} = k$
Now, In xy plane $z = 0$
$k = - \frac{1}{5}$
$∴ \frac{x - 3}{2} = \frac{- 1}{5} and \frac{y - 4}{- 3} = - \frac{1}{5}$
$x - 3 = - \frac{2}{5} and y - 4 = 3 / 5$
$x = - \frac{2}{5} + 3, y = \frac{3}{5} + 4$
$x = \frac{13}{5} and y = \frac{23}{5}$
Hence, the required co-ordinate $(\frac{13}{5}, \frac{23}{5}, 0)$

[JCECE-2018]

Co-Ordinate system

88284 If $y = f (x)$ makes positive intercepts of 2 and 1 unit an $x$ and $y$ coordinate axes and encloses an area of $\frac{3}{4}$ square unit with the axes, then $\int_{0}^{2} x f^{'} (x) d x$ is is

1

\frac{3}{2}

2 1

3

\frac{5}{4}

4

- \frac{3}{4}

Explanation:

(D) : Given,
$\int_{0}^{2} f (x) d x = 3 / 4$
$∴ \int_{0}^{2} x f^{'} (x) d x$
$= x \int_{0}^{2} f^{'} (x) d x - \int_{0}^{2} f (x) d x$

$= {[x \int_{0}^{2} f (x) dx]}_{0}^{2} - 3 / 4 = 2 f (2) - 3 / 4$
$= 2 \times 0 - 3 / 4$
$[y = f (x) \Rightarrow y = f (2)] = - 3 / 4$

BCECE-2018

Co-Ordinate system

88285 If $A (2, 3)$ and $B (- 2, 1)$ are two vertices of a triangle and third vertex moves on the line $2 x + 3 y = 9$ , then the locus of the centroid of the triangle is

1

2 x + 3 y = 1

2

2 x + y = 3

3

2 x - 3 y = 1

4

x - y = 1

Explanation:

(A) : Given,
The co-ordinate of vertices of triangle are $A (2, - 3), B (-$ $2, 1)$ and $C (x, y)$
Now, Let the co-ordinate of centroid be $(h, k)$
$∴ h = (\frac{2 - 2 + x}{3}) \Rightarrow x = 3 h$
$k = (\frac{- 3 + 1 + y}{3}) \Rightarrow y = 3 k + 2$
Where $(x, y)$ are the point on the line $2 x + 3 y = 9$
$∴ 2 (3 h) + 3 (3 k + 2) = 9$
$6 h + 9 k + 6 = 9$
$6 h + 9 k = 3$
$2 h + 3 k = 1$
Locus of the centroid of the triangle is $2 x + 3 y = 1$

BCECE-2017

Co-Ordinate system

88287 Let $A (2, - 3)$ and $B (- 2, 1)$ be vertices of a $△ ABC$ . If The centroid of this triangle moves on the line $2 x + 3 y = 1$ , then the locus of the vertex $C$ is the line

1

2 x + 3 y = 9

2

2 x - 3 y = 7

3

3 x + 2 y = 5

4

3 x - 2 y = 3

Explanation:

(A) : Given,
The co-ordinate of vertices of triangle $A (2, - 3), B (- 2, 1)$ and $C (h, k)$
Now, co-ordinate of centroid lies on the line $2 x + 3 y = 1$

$\Rightarrow (\frac{2 - 2 + h}{3}, \frac{- 3 + 1 + k}{3}) \Rightarrow (\frac{h}{3}, \frac{- 2 + k}{3})$
$∴ (\frac{h}{3}, \frac{- 2 + k}{3})$ satisfy the equation ${2 x + 3 y = 1}$
$\Rightarrow 2 (\frac{h}{3}) + 3 (\frac{- 2 + k}{3}) = 1$
$\frac{2}{3} h + (- 2 + k) = 1$
$\Rightarrow 2 h - 6 + 3 k = 3$
$2 h + 3 k = 9$
Locus of the vertex $C$ is the line $2 x + 3 y = 9$

BCECE-2014

Co-Ordinate system

88283 The coordinates of the point on the line through the points $A (3, 4, 1)$ and $B (5, 1, 6)$ crosses XY-plane are

1

(\frac{13}{5}, \frac{23}{5}, 0)

2

(\frac{3}{5}, \frac{2}{5}, 0)

3

(1, 1, 0)

4

(- \frac{13}{5}, \frac{23}{5}, 0)

Explanation:

(A) : Given,
The co-ordinate of the points $A (3, 4, 1)$ and $B (5, 1, 6)$
Now,
The equation of line in space can be written as
$\frac{x - x_{1}}{x_{2} - x_{1}} = \frac{y - y_{1}}{y_{2} - y_{1}} = \frac{z - z_{1}}{z_{2} - z_{1}} [\begin{array}{l} Where (x_{1}, y_{1}, z_{1}) and \\ (x_{2}, y_{2}, z_{2}) are the points \end{array}]$
$\frac{x - 3}{5 - 3} = \frac{y - 4}{1 - 4} = \frac{z - 1}{6 - 1}$
$\frac{x - 3}{2} = \frac{y - 4}{- 3} = \frac{z - 1}{5} = k$
Now, In xy plane $z = 0$
$k = - \frac{1}{5}$
$∴ \frac{x - 3}{2} = \frac{- 1}{5} and \frac{y - 4}{- 3} = - \frac{1}{5}$
$x - 3 = - \frac{2}{5} and y - 4 = 3 / 5$
$x = - \frac{2}{5} + 3, y = \frac{3}{5} + 4$
$x = \frac{13}{5} and y = \frac{23}{5}$
Hence, the required co-ordinate $(\frac{13}{5}, \frac{23}{5}, 0)$

[JCECE-2018]

Co-Ordinate system

88284 If $y = f (x)$ makes positive intercepts of 2 and 1 unit an $x$ and $y$ coordinate axes and encloses an area of $\frac{3}{4}$ square unit with the axes, then $\int_{0}^{2} x f^{'} (x) d x$ is is

1

\frac{3}{2}

2 1

3

\frac{5}{4}

4

- \frac{3}{4}

Explanation:

(D) : Given,
$\int_{0}^{2} f (x) d x = 3 / 4$
$∴ \int_{0}^{2} x f^{'} (x) d x$
$= x \int_{0}^{2} f^{'} (x) d x - \int_{0}^{2} f (x) d x$

$= {[x \int_{0}^{2} f (x) dx]}_{0}^{2} - 3 / 4 = 2 f (2) - 3 / 4$
$= 2 \times 0 - 3 / 4$
$[y = f (x) \Rightarrow y = f (2)] = - 3 / 4$

BCECE-2018

Co-Ordinate system

88285 If $A (2, 3)$ and $B (- 2, 1)$ are two vertices of a triangle and third vertex moves on the line $2 x + 3 y = 9$ , then the locus of the centroid of the triangle is

1

2 x + 3 y = 1

2

2 x + y = 3

3

2 x - 3 y = 1

4

x - y = 1

Explanation:

(A) : Given,
The co-ordinate of vertices of triangle are $A (2, - 3), B (-$ $2, 1)$ and $C (x, y)$
Now, Let the co-ordinate of centroid be $(h, k)$
$∴ h = (\frac{2 - 2 + x}{3}) \Rightarrow x = 3 h$
$k = (\frac{- 3 + 1 + y}{3}) \Rightarrow y = 3 k + 2$
Where $(x, y)$ are the point on the line $2 x + 3 y = 9$
$∴ 2 (3 h) + 3 (3 k + 2) = 9$
$6 h + 9 k + 6 = 9$
$6 h + 9 k = 3$
$2 h + 3 k = 1$
Locus of the centroid of the triangle is $2 x + 3 y = 1$

BCECE-2017

Co-Ordinate system

88287 Let $A (2, - 3)$ and $B (- 2, 1)$ be vertices of a $△ ABC$ . If The centroid of this triangle moves on the line $2 x + 3 y = 1$ , then the locus of the vertex $C$ is the line

1

2 x + 3 y = 9

2

2 x - 3 y = 7

3

3 x + 2 y = 5

4

3 x - 2 y = 3

Explanation:

(A) : Given,
The co-ordinate of vertices of triangle $A (2, - 3), B (- 2, 1)$ and $C (h, k)$
Now, co-ordinate of centroid lies on the line $2 x + 3 y = 1$

$\Rightarrow (\frac{2 - 2 + h}{3}, \frac{- 3 + 1 + k}{3}) \Rightarrow (\frac{h}{3}, \frac{- 2 + k}{3})$
$∴ (\frac{h}{3}, \frac{- 2 + k}{3})$ satisfy the equation ${2 x + 3 y = 1}$
$\Rightarrow 2 (\frac{h}{3}) + 3 (\frac{- 2 + k}{3}) = 1$
$\frac{2}{3} h + (- 2 + k) = 1$
$\Rightarrow 2 h - 6 + 3 k = 3$
$2 h + 3 k = 9$
Locus of the vertex $C$ is the line $2 x + 3 y = 9$

BCECE-2014