If $(x, y)$ be on the line joining the two points $(1, -3)$ and $(-4, 2)$, prove that $x + y + 2 = 0$.
Given:
$(x, y)$ is on the line joining the two points $(1, -3)$ and $(-4, 2)$.
To do:
We have to prove that $x + y + 2 = 0$.
Solution:
We know that,
If the points $A, B$ and $C$ are collinear then the area of $\triangle ABC$ is zero.
Let $A(1, -3), B(x, y)$ and $C(-4, 2)$ be the vertices of $\triangle ABC$.
Area of a triangle with vertices $(x_1,y_1), (x_2,y_2), (x_3,y_3)$ is given by,
Area of $\Delta=\frac{1}{2}[x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})]$
Therefore,
Area of triangle \( ABC=\frac{1}{2}[1(y-2)+x(2+3)+(-4)(-3-y)] \)
\( 0=\frac{1}{2}[y-2+x(5)-(-4)(3+y)] \)
\( 0(2)=(y-2+5x+12+4y) \)
\( 0=5x+5y+10 \)
\( 5(x+y+2)=0 \)
\( x+y+2=0 \)
Hence proved.
Related Articles
- If the point $C (-1, 2)$ divides internally the line segment joining the points $A (2, 5)$ and $B (x, y)$ in the ratio $3 : 4$, find the value of $x^2 + y^2$.
- If the points $(-2, -1), (1, 0), (x, 3)$ and $(1, y)$ form a parallelogram, find the values of $x$ and $y$.
- 1. Factorize the expression \( 3 x y - 2 + 3 y - 2 x \)A) \( (x+1),(3 y-2) \)B) \( (x+1),(3 y+2) \)C) \( (x-1),(3 y-2) \)D) \( (x-1),(3 y+2) \)2. Factorize the expression \( \mathrm{xy}-\mathrm{x}-\mathrm{y}+1 \)A) \( (x-1),(y+1) \)B) \( (x+1),(y-1) \)C) \( (x-1),(y-1) \)D) \( (x+1),(y+1) \)
- A pair of linear equations which has a unique solution \( x=2, y=-3 \) is(A) \( x+y=-1 \)\( 2 x-3 y=-5 \)(B) \( 2 x+5 y=-11 \)\( 4 x+10 y=-22 \)(C) \( 2 x-y=1 \)\( 3 x+2 y=0 \)(D) \( x-4 y-14=0 \)\( 5 x-y-13=0 \)
- If the points $(2, 1)$ and $(1, -2)$ are equidistant from the point $(x, y)$, show that $x + 3y = 0$.
- \Find $(x +y) \div (x - y)$. if,(i) \( x=\frac{2}{3}, y=\frac{3}{2} \)(ii) \( x=\frac{2}{5}, y=\frac{1}{2} \)(iii) \( x=\frac{5}{4}, y=\frac{-1}{3} \)(iv) \( x=\frac{2}{7}, y=\frac{4}{3} \)(v) \( x=\frac{1}{4}, y=\frac{3}{2} \)
- Verify that \( x^{3}+y^{3}+z^{3}-3 x y z=\frac{1}{2}(x+y+z)\left[(x-y)^{2}+(y-z)^{2}+(z-x)^{2}\right] \)
- Solve the following pairs of equations by reducing them to a pair of linear equations:(i) \( \frac{1}{2 x}+\frac{1}{3 y}=2 \)\( \frac{1}{3 x}+\frac{1}{2 y}=\frac{13}{6} \)(ii) \( \frac{2}{\sqrt{x}}+\frac{3}{\sqrt{y}}=2 \)\( \frac{4}{\sqrt{x}}-\frac{9}{\sqrt{y}}=-1 \)(iii) \( \frac{4}{x}+3 y=14 \)\( \frac{3}{x}-4 y=23 \)(iv) \( \frac{5}{x-1}+\frac{1}{y-2}=2 \)\( \frac{6}{x-1}-\frac{3}{y-2}=1 \)(v) \( \frac{7 x-2 y}{x y}=5 \)\( \frac{8 x+7 y}{x y}=15 \),b>(vi) \( 6 x+3 y=6 x y \)\( 2 x+4 y=5 x y \)4(vii) \( \frac{10}{x+y}+\frac{2}{x-y}=4 \)\( \frac{15}{x+y}-\frac{5}{x-y}=-2 \)(viii) \( \frac{1}{3 x+y}+\frac{1}{3 x-y}=\frac{3}{4} \)\( \frac{1}{2(3 x+y)}-\frac{1}{2(3 x-y)}=\frac{-1}{8} \).
- Find $25 x^{2}+16 y^{2}$, if $5 x+4 y=8$ and $x y=1$.
- Simplify: \( \frac{11}{2} x^{2} y-\frac{9}{4} x y^{2}+\frac{1}{4} x y-\frac{1}{14} y^{2} x+\frac{1}{15} y x^{2}+\frac{1}{2} x y \).
- If points $( a,\ 0),\ ( 0,\ b)$ and $( x,\ y)$ are collinear, prove that $\frac{x}{a}+\frac{y}{b}=1$.
- Find a relation between $x$ and $y$, if the points $(x, y), (1, 2)$ and $(7, 0)$ are collinear.
- If \( 2^{x}=3^{y}=6^{-z} \), show that \( \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0 \).
- Simplify: $\frac{x^{-3}-y^{-3}}{x^{-3} y^{-1}+(x y)^{-2}+y^{-1} x^{-3}}$.
- Check if the following is true.\( (x+y)^{2}=(x-y)^{2}+4 x y \)
Kickstart Your Career
Get certified by completing the course
Get Started
Data Structure
Networking
RDBMS
Operating System
Java
MS Excel
iOS
HTML
CSS
Android
Python
C Programming
C++
C#
MongoDB
MySQL
Javascript
PHP
Physics
Chemistry
Biology
Mathematics
English
Economics
Psychology
Social Studies
Fashion Studies
Legal Studies