5 Marks Questions

Question 15 Marks

If $a \neq b \neq 0$, prove that the points $\left(a, a^2\right),\left(b, b^2\right)$, $(0,0)$ will not be collinear.

Answer

We know that three points are collinear if the area of the triangle formed by them is zero.
The area of triangle can be calculated using the formula
$A=\frac{1}{2}\left|x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right|$
Here, $x_1=a, y_1=a^2, x_2=b, y_2=b^2, x_3=0, y_3=0$
$A=\frac{1}{2}\left|a\left(b^2-0\right)+b\left(0-a^2\right)+0\left(a^2-b^2\right)\right|$
$A=\frac{1}{2}\left|a b^2-a^2 b\right|$
According to the question $a \neq b \neq 0$
$\therefore A=\frac{1}{2}\left|a b^2-a^2 b\right| \neq 0$
Hence, the points $\left(a, a^2\right),\left(b, b^2\right),(0,0)$ are non collinear.
Hence proved.

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Question 25 Marks

If the points $A(k+1,2 k), B(3 k, 2 k+3)$ and $C(5 k-1,5 k)$ are collinear, then find the value of $k$.

Answer

Given points are collinear so area of triangle formed must be zero.
$\Rightarrow \frac{1}{2}\left|x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right|=0$
Given;
$\begin{array}{l}\Rightarrow x_1=k+1, y_1=2 k ; \\\Rightarrow x_2=3 k, y_2=2 k+3; \\\Rightarrow x_3=5 k-1, y_3=5 k ; \\\Rightarrow \frac{1}{2}\left|\begin{array}{l}(k+1)\{(2 k+3)-(5 k)\}+(3 k)\{5 k-2 k\} \\+(5 k-1)\{(2 k)-(2 k+3)\}\end{array}\right|=0\end{array}$
$\begin{array}{l}\Rightarrow(k+1)\{-3 k+3\}+(3 k)\{3 k\}+(5 k-1)\{-3\}=0 \\\Rightarrow-3 k^2+3 k-3 k+3+9 k^2-15 k+3=0 \\\Rightarrow 6 k^2-15 k+6=0\end{array}$
Dividing the above equation by 3
$\begin{array}{l}\Rightarrow 2 k^2-5 k+2=0 \\\Rightarrow 2 k^2-4 k-k+2=0 \\\Rightarrow 2 k(k-2)-1(k-2)=0 \\\Rightarrow(k-2)(2 k-1)=0 \\\Rightarrow k-2=0 \text { and } 2 k-1=0 \\\Rightarrow k=2 \text { and } k=\frac{1}{2}\end{array}$
Therefore, the values of $k=\frac{1}{2}$ and $2$

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Question 35 Marks

In fig., the vertices of $\triangle A B C$ are $A(4,6), B(1,5)$, $C(7,2) . A$ line segment $D E$ is drawn to intersect the sides $A B$ and $A C$ at $D$ and $E$ respectively such that $\frac{A D}{A B}=\frac{A E}{A C}=\frac{1}{3}$. Calculate the area of and compare it with area of $\triangle A B C$.

Answer

Given $\frac{A D}{A B}=\frac{A E}{A C}=\frac{1}{3}$
$\begin{array}{l}\frac{A D}{A B}=\frac{1}{3} \text { and } \frac{A E}{A C}=\frac{1}{3} \\\frac{AD}{AB}=\frac{1}{3} \\\Rightarrow 3 AD=AD+DB \Rightarrow 3 AD-AD=DB \\\Rightarrow 2 AD=BD \Rightarrow \frac{AD}{BD}=\frac{1}{2}\end{array}$
Thus point $D$ divides line AB in ratio 1:2

Using section formula, Coordinates of $D$
$\left(\frac{m_1 x_2+m_2 x_1}{m_1+m_2}, \frac{m_1 y_2+m_2 y_1}{m_1+m_2}\right)$
Putting values in the given above formula
$\begin{array}{l}m_1=1, m_2=2 \\x_1=4, x_2=1 \\y_1=6, y_2=5\end{array}$
Point $D=\left(\frac{m_1 x_2+m_2 x_1}{m_1+m_2}, \frac{m_1 y_2+m_2 y_1}{m_1+m_2}\right)$
$=\left(\frac{1 \times 1+2 \times 4}{1+2}, \frac{1 \times 5+2 \times 6}{1+2}\right)=\left(\frac{1+8}{3}, \frac{5+12}{3}\right)$
$\begin{array}{l}\text{Point D}=\left(\frac{9}{3}, \frac{17}{3}\right) \\\frac{AE}{AC}=\frac{1}{3} \\\Rightarrow 3 AE=AE+CE \Rightarrow 3 AE-AE=CE \\\Rightarrow 2 AE=CE \Rightarrow \frac{AE}{CE}=\frac{1}{2}\end{array}$
Thus point $E$ divides line $A C$ in ratio $1: 2$

Using section formula, Coordinates of $E$
$\left(\frac{m_1 x_2+m_2 x_1}{m_1+m_2}, \frac{m_1 y_2+m_2 y_1}{m_1+m_2}\right)$
Putting values in the given above formula
$\begin{array}{l}m_1=1, m_2=2 \\x_1=4, x_2=7 \\y_1=6, y_2=2 \\\text { Point } E=\left(\frac{m_1 x_2+m_2 x_1}{m_1+m_2}, \frac{m_1 y_2+m_2 y_1}{m_1+m_2}\right) \\=\left(\frac{1 \times 7+2 \times 4}{1+2}, \frac{1 \times 2+2 \times 6}{1+2}\right)=\left(\frac{7+8}{3}, \frac{2+12}{3}\right) \\=\left(\frac{15}{3}, \frac{14}{3}\right)\end{array}$
Now, Finding area of $\triangle A B C$ and $\triangle A D E$

Area of $\triangle A B C$
$=\frac{1}{2}\left[x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right]$
Given,
$\begin{array}{c}x_1=4, x_2=1 \\x_3=7, y_1=6 \\y_2=5, y_3=2\end{array}$
$\begin{array}{l}=\frac{1}{2}\left[x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right] \\=\frac{1}{2}[4(5-2)+1(2-6)+7(6-5)] \\=\frac{1}{2}[4(3)+1(-4)+7(1)] \\=\frac{1}{2}[12-4+7] \\=\frac{15}{2} \text { sq.units }\end{array}$
Area of $\triangle A D E$
$=\frac{1}{2}\left[x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right]$
Given,
$\begin{array}{l}x_1=4, x_2=\frac{9}{3} \\x_3=\frac{15}{3}, y_1=6 \\y_2=\frac{17}{3}, y_3=\frac{14}{3} \\=\frac{1}{2}\left[4\left(\frac{17}{3}-\frac{14}{3}\right)+\frac{9}{3}\left(\frac{14}{3}-6\right)+\frac{15}{3}\left(6-\frac{17}{3}\right)\right] \\
=\frac{1}{2}\left[4(1)+3\left(\frac{-4}{3}\right)+5\left(\frac{1}{3}\right)\right] \\=\frac{1}{2} \times \frac{5}{3}=\frac{5}{6} \text { sq units }\end{array}$
Therefore,
$\frac{\operatorname{ar}(\triangle ADE)}{\operatorname{ar}(\triangle ABC)}=\frac{\frac{5}{6}}{\frac{15}{2}}=\frac{5}{6} \times\frac{2}{15}=\frac{1}{9}$
Hence, ratio is $1: 9$.

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Question 45 Marks

Find the values of so that the area of the triangle with vertices $(1,-1),(-4,2 k)$ and $(-k,-5)$ is $24$ square units.

Answer

Area of a triangle with coordinates $\left(x_1, y_1\right)$, $\left(x_2, y_2\right),\left(x_3, y_3\right)$ is given by-
$=\frac{1}{2}\left|x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right|$
Putting in the coordinates,
$\begin{array}{l}x_1=1, x_2=-4, x_3=-k, y_1=-1, y_2=2 k, y_3=-5 \\\Rightarrow \frac{1}{2}[1(2 k+5)-4(-5+1)-k(-1-2 k)]=24 \\\Rightarrow \frac{1}{2}\left[2 k+5+16+k+2 k^2\right]=24 \\\Rightarrow \frac{1}{2}\left[2 k^2+3 k+21\right]=24 \Rightarrow 2 k^2+3 k+21=48 \\\Rightarrow 2 k^2+3 k-27=0 \\\Rightarrow 2 k^2-6 k+9 k-27=0 \Rightarrow 2 k(k-3)+9(k-3)=0 \\\Rightarrow(k-3)(2 k+9)=0 \Rightarrow k-3=0 \text { and } 2 k+9=0 \\\Rightarrow k=3 \text { and } k=\frac{-9}{2}\end{array}$

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Question 55 Marks

If $A(-4,8), B(-3,-4), C(0,-5)$ and $D(5,6)$ are the vertices of a quadrilateral $A B C D$, find its area.

Answer

In this quadrilateral $A B C D$, we have 2 triangles i.e. $\triangle A B C$ and $\triangle A C D$
Area of a triangle
$\begin{array}{l}=\frac{1}{2}\left\{x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right\} \\\Rightarrow \text { Area of } \triangle ABC=\frac{1}{2}\left\{\begin{array}{l}-4(-4-(-5))-3(-5-8) +0(8-(-4))\end{array}\right\}\end{array}$
Now,
$\begin{array}{l}\Rightarrow \text { Area of } \triangle ABC=\frac{1}{2}\left\{\begin{array}{l}-4(-4-(-5))-3(-5-8) +0(8-(-4))\end{array}\right\} \\=\frac{1}{2}\{-4(-4+5)-3(-13)+0\} \\=\frac{1}{2}\{-4+39\}=\frac{35}{2}=17.5\end{array}$
Area of $\triangle ABC =17.5$ units $^2$
Now,
$\begin{array}{l}\text { Area of } \triangle ACD=\frac{1}{2}\{-4(-5-6)+0(6-8)+5(8-(-5))\} \\=\frac{1}{2}\{-4(-11)+0+5(8+5)\} \\=\frac{1}{2}\{44+5(13)\}=\frac{1}{2}\{44+65\} \\=\frac{109}{2}=54.5\end{array}$
Area of $\triangle ACD =54.5$ units $^2$
Now area of quadrilateral $A B C D$ will be sum of area of $\triangle A B C$ and $\triangle A C D$.
Therefore Area of quadrilateral $A B C D=17.5+$ $54.5=72$ sq. units
Hence, answer is 72 sq. units.

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Question 65 Marks

If $A(-3,5), B(-2,-7), C(1,-8)$ and $D(6,3)$ are the vertices of a quadrilateral $A B C D$, find its area.

Answer

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Question 75 Marks

Find the ratio in which the point $P(x, 2)$ divides the line segment joining the points $A(12,5)$ and $B(4,-3)$. Also find the value of $x$.

Answer

Let us assume the point $P$ divides the line segment joining the points $A(12,5)$ and $B(4,-3)$ in the ratio $r: 1$
Using section formula $\left(\frac{m x_2+x_1}{m+1}, \frac{m y_2+y_1}{m+1}\right)$
Then the coordinates of $P$ are, $\left(\frac{4 r+12}{r+1}, \frac{-3 r+5}{r+1}\right)$
Given, the coordinates of $P$ are $(x, 2)$
$\begin{array}{l}\therefore(x, 2)=\left(\frac{4 r+12}{r+1}, \frac{-3 r+5}{r+1}\right) \\\therefore \frac{4 r+12}{r+1}=x \text { and } \frac{-3 r+5}{r+1}=2\end{array}$
Now, $\frac{-3 r+5}{r+1}=2$
$\begin{array}{l}\Rightarrow-3 r+5=2(r+1) \\\Rightarrow 5 r=3 \Rightarrow r=\frac{3}{5}\end{array}$
Substitute $r=\frac{3}{5}$ in $\frac{4 r+12}{r+1}=x$, we get
$\begin{array}{l}\Rightarrow x=\frac{4\left(\frac{3}{5}\right)+12}{\frac{3}{5}+1} \\\Rightarrow x=\frac{\frac{12}{5}+12}{\frac{3}{5}+1} \\\Rightarrow x=\frac{\frac{72}{5}}{\frac{8}{5}} \Rightarrow x=\frac{72}{8} \\\Rightarrow x=9\end{array}$
Hence, the value of $x$ is 9 and the point $P(x, 2)$ divides the line segment joining the points $A(12,5)$ and $B(4,-3)$ in the ratio $\frac{3}{5}: 1=3: 5$

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Question 85 Marks

The three vertices of a parallelogram $A B C D$ are $A(3,-4), B(-1,-3)$ and $C(-6,2)$. Find the coordinates of vertex $D$ and find the area of parallelogram $A B C D$.

Answer

The three vertices are $A(3,-4), B(-1,-3)$ and $C(-6,2)$.
Let the coordinates of the vertex $D$ be $(x, y)$.
In a parallelogram, the diagonals bisect each other.
Hence, Mid point of $A C=$ Mid point of $B D$
$\begin{array}{l}\Rightarrow\left(\frac{3+(-6)}{2}, \frac{-4+2}{2}\right)=\left(\frac{-1+x_2}{2}, \frac{-3+y_2}{2}\right) \\\Rightarrow\left(\frac{-3}{2}, \frac{-2}{2}\right)=\left(\frac{-1+x_2}{2}, \frac{-3+y_2}{2}\right) \\\Rightarrow-3=-1+x_2,-2=-3+y_2 \\\Rightarrow x=-2, y=1 .\end{array}$
Hence, the coordinates of the fourth vertex $D$ is $(-2,1)$.
Now, for the area of $A B C D=$ area of triangle $A B C$ + area of triangle $A C D$
$=2 \times$ area of triangle $A B C$
The area of a triangle with vertices $\left(x_1, y_1\right),\left(x_2, y_2\right)$ and $\left(x_3, y_3\right)$ is
$\frac{1}{2}\left[x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right]$
Now substituting the values,
$\begin{array}{l}\text {Area} =\frac{1}{2}|[3(-3-2)+(-1)(2-(-4))+(-6)(-4-(-3))]| \\=\frac{1}{2}|[-15-6+6]| \\=\frac{|-15|}{2} \text { square units }\end{array}$
Hence area of parallelogram $2 \times \frac{15}{2}=15$ square units.

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