Question 1

Find area of the triangle with vertices at the point given in each of the following:

(i) $(1,0),(6,0),(4,3)$

(ii) $(2,7),(1,1),(10,8)$

(iii) $(-2,-3),(3,2),(-1,-8)$

Accepted Answer

(i) The area of the triangle with vertices $(1,0),(6,0),(4,3)$ is given by the relation,

$$

\begin{aligned}

\Delta & =\frac{1}{2}\left|\begin{array}{lll}

1 & 0 & 1 \

6 & 0 & 1 \

4 & 3 & 1

\end{array}ight| \

& =\frac{1}{2}[1(0-3)-0(6-4)+1(18-0)] \

& =\frac{1}{2}[-3+18] \

& =\frac{1}{2}[15] \

& =\frac{15}{2}

\end{aligned}

$$

Hence, area of the triangle is $\frac{15}{2}$ square units.

(ii) The area of the triangle with vertices $(2,7),(1,1),(10,8)$ is given by the relation,

$$

\begin{aligned}

\Delta & =\frac{1}{2}\left|\begin{array}{ccc}

2 & 7 & 1 \

1 & 1 & 1 \

10 & 8 & 1

\end{array}ight| \

& =\frac{1}{2}[2(1-8)-7(1-10)+1(8-10)] \

& =\frac{1}{2}[2(-7)-7(-9)+1(-2)] \

& =\frac{1}{2}[-14+63-2] \

& =\frac{1}{2}[47] \

& =\frac{47}{2}

\end{aligned}

$$

Hence, area of the triangle is $\frac{47}{2}$ square units.

(iii) The area of the triangle with vertices $(-2,-3),(3,2),(-1,-8)$ is given by the relation,

$$

\begin{aligned}

\Delta & =\frac{1}{2}\left|\begin{array}{ccc}

-2 & -3 & 1 \

3 & 2 & 1 \

-1 & -8 & 1

\end{array}ight| \

& =\frac{1}{2}[-2(2+8)+3(3+1)+1(-24+2)] \

& =\frac{1}{2}[-2(10)+3(4)+1(-22)] \

& =\frac{1}{2}[-20+12-22] \

& =-\frac{1}{2}[30] \

& =-15

\end{aligned}

$$

Hence, area of the triangle is 15 square units.

Question 2

Show that the points $A(a, b+c), B(b, c+a), C(c, a+b)$ are collinear.

Accepted Answer

The area of the triangle with vertices $A(a, b+c), B(b, c+a), C(c, a+b)$ is given by the absolute value of the relation:

$$

\begin{array}{rlrl}

\Delta & =\frac{1}{2}\left|\begin{array}{lll}

a & b+c & 1 \

b & c+a & 1 \

c & a+b & 1

\end{array}ight| & \

& =\frac{1}{2}\left|\begin{array}{ccc}

a & b+c & 1 \

b-a & a-b & 0 \

c-a & a-c & 0

\end{array}ight| & & \

& =\frac{1}{2}(a-b)(c-a)\left|\begin{array}{ccc}

a & b+c & 1 \

-1 & 1 & 0 \

1 & -1 & 0

\end{array}ight| & & \

& =\frac{1}{2}(a-b)(c-a)\left|\begin{array}{ccc}

a & b+c & 1 \

-1 & 1 & 0 \

0 & 0 & 0

\end{array}ight| & & {\left[R_{3} ightarrow R_{2}-R_{1} 	ext { and } R_{3} ightarrow R_{3}-R_{1}ight]} \

& =0 & &

\end{array}

$$

Thus, the area of the triangle formed by points is zero.

Hence, the points are collinear.

Question 3

Find values of $k$ if area of triangle is 4 square units and vertices are:

(i) $(k, 0),(4,0),(0,2)$

(ii) $(-2,0),(0,4),(0, k)$

Accepted Answer

We know that the area of a triangle whose vertices are $\left(x_{1}, y_{1}ight),\left(x_{2}, y_{2}ight)$ and $\left(x_{3}, y_{3}ight)$ is the absolute value of the determinant $(\Delta)$, where

$$

\Delta=\frac{1}{2}\left|\begin{array}{lll}

x_{1} & y_{1} & 1 \

x_{2} & y_{2} & 1 \

x_{3} & y_{3} & 1

\end{array}ight|

$$

It is given that the area of triangle is 4 square units.

Hence, $\Delta= \pm 4$

(i) The area of the triangle with vertices $(k, 0),(4,0),(0,2)$ is given by the relation,

$$

\begin{aligned}

\Delta & =\frac{1}{2}\left|\begin{array}{lll}

k & 0 & 1 \

4 & 0 & 1 \

0 & 2 & 1

\end{array}ight| \

& =\frac{1}{2}[k(0-2)-0(4-0)+1(8-0)] \

& =\frac{1}{2}[-2 k+8] \

& =-k+4

\end{aligned}

$$

Therefore, $-k+4= \pm 4$

When $-k+4=-4$

Then $k=8$

When $-k+4=4$

Then $k=0$

Hence, $k=0,8$

(ii) The area of the triangle with vertices $(-2,0),(0,4),(0, k)$ is given by the relation,

$$

\begin{aligned}

\Delta & =\frac{1}{2}\left|\begin{array}{ccc}

-2 & 0 & 1 \

0 & 4 & 1 \

0 & k & 1

\end{array}ight| \

& =\frac{1}{2}[-2(4-k)] \

& =k-4

\end{aligned}

$$

Therefore, $-k+4= \pm 4$

When $k-4=4$

Then $k=8$

When $k-4=-4$

Then $k=0$

Hence, $k=0,8$

Question 4

(i) Find equation of line joining $(1,2)$ and $(3,6)$ using determinants.

(ii) Find equation of line joining $(3,1)$ and $(9,3)$ using determinants.

Accepted Answer

(i) Let $P(x, y)$ be any point on the line joining points $A(1,2)$ and $B(3,6)$. Then, the points $A, B$ and $P$ are collinear.

Hence, the area of triangle $A B P$ will be zero.

Therefore,

$$

\begin{aligned}

& \Rightarrow \frac{1}{2}\left|\begin{array}{lll}

1 & 2 & 1 \

3 & 6 & 1 \

x & y & 1

\end{array}ight|=0 \

& \Rightarrow \frac{1}{2}[1(6-y)-2(3-x)+1(3 y-6 x)]=0 \

& \Rightarrow 6-y-6+2 x+3 y-6 x=0 \

& \Rightarrow 2 y-4 x=0 \

& \Rightarrow y=2 x

\end{aligned}

$$

Thus, the equation of the line joining the given points is $y=2 x$.

(ii) Let $P(x, y)$ be any point on the line joining points $A(3,1)$ and $B(9,3)$.

Then, the points $A, B$ and $P$ are collinear.

Hence, the area of triangle $A B P$ will be zero.

Therefore,

$$

\begin{aligned}

& \Rightarrow \frac{1}{2}\left|\begin{array}{lll}

3 & 1 & 1 \

9 & 3 & 1 \

x & y & 1

\end{array}ight|=0 \

& \Rightarrow \frac{1}{2}[3(3-y)-1(9-x)+1(9 y-3 x)]=0 \

& \Rightarrow 9-3 y-9+x+9 y-3 x=0 \

& \Rightarrow 6 y-2 x=0 \

& \Rightarrow x-3 y=0

\end{aligned}

$$

Thus, the equation of the line joining the given points is $x-3 y=0$.

Question 5

If area of the triangle is 35 square units with vertices $(2,-6),(5,4),(k, 4)$. Then $k$ is

(A) 12

(B) -2

(C) $-12,-2$

(D) $12,-2$

Accepted Answer

The area of the triangle with vertices $(2,-6),(5,4),(k, 4)$ is given by the relation,

$$

\begin{aligned}

\Delta & =\frac{1}{2}\left|\begin{array}{ccc}

2 & -6 & 1 \

5 & 4 & 1 \

k & 4 & 1

\end{array}ight| \

& =\frac{1}{2}[2(4-4)+6(5-k)+1(20-4 k)] \

& =\frac{1}{2}[30-6 k+20-4 k] \

& =\frac{1}{2}[50-10 k] \

& =25-5 k

\end{aligned}

$$

It is given that the area of the triangle is 35 square units

Hence, $\Delta= \pm 35$.

Therefore,

$$

\begin{aligned}

& \Rightarrow 25-5 k= \pm 35 \

& \Rightarrow 5(5-k)= \pm 35 \

& \Rightarrow 5-k= \pm 7

\end{aligned}

$$

When, $5-k=-7$

Then, $k=12$

When, $5-k=7$

Then, $k=-2$

Hence, $k=12,-2$

Thus, the correct option is D.

\section*{EXERCISE 4.4}