Question

In the graph of the linear equation $5x + 2y = 110$ there is a point such that its ordinate is one fourth of abscissa. Find coordinates of the point :

Answer 1

Hint:
We should know that abscissa is the x-coordinate of a point, whereas ordinate is the y-coordinate of a point. According to the question a relation is given between ordinate and abscissa. We will substitute the relation in the given equation $5x + 2y = 110$ to get the final answer.

Complete step by step solution:
According to the question, the linear equation is
 $5x + 2y = 110$ … (1)

We know that,
 \[Abscissa \to \,x\,coordinate\,of\,the\,point\]
 \[Ordinate \to \,y\,coordinate\,of\,the\,point\]

Let us assume the required point as \[\left( {x,y} \right)\]
So, here x is the abscissa of the point and y is the ordinate of the point.

According to the question, we get
 $ \Rightarrow y = \dfrac{x}{4}$ … (2)

Put $y = \dfrac{x}{4}$ in (1), we get
 $ \Rightarrow 5x + \dfrac{{2x}}{4} = 110$

On simplification we get,
 $ \Rightarrow 5x + \dfrac{x}{2} = 110$

Taking LCM at LHS, we get
 $ \Rightarrow \dfrac{{10x + x}}{2} = 110$

On simplification we get,
$ \Rightarrow \dfrac{{11x}}{2} = 110$

By cross multiplying, we get
 $ \Rightarrow 11x = 220$

On dividing the equation by 11 we get,
 $ \Rightarrow x = \dfrac{{220}}{{11}}$

On simplification we get,
 $ \Rightarrow x = 20$ … (3)

Put $x = 20$ in (1), we get
 $ \Rightarrow 5 \times 20 + 2y = 110$

On multiplication of first term we get,
 $ \Rightarrow 100 + 2y = 110$

Subtracting 100 from both sides we get,
 $ \Rightarrow 2y = 110 - 100$

On simplification we get,
 $ \Rightarrow 2y = 10$

On dividing the equation by 2 we get,
 $ \Rightarrow y = \dfrac{{10}}{2}$

Hence we have,
 $ \Rightarrow y = 5$ … (4)

Hence, in the required point \[\left( {x,y} \right)\]
 $x = 20$
 $ \Rightarrow y = 5$

Hence, the coordinate of the required point is \[\left( {20,5} \right)\]

Note:
The above question was done using the substitution method, there is another alternate way using Elimination. According to the question,
 $5x + 2y = 110$ … (1)
Let us assume the required point as \[\left( {x,y} \right)\]
So, here x is the abscissa of the point and y is the ordinate of the point.

According to the question, we get
 $ \Rightarrow y = \dfrac{x}{4}$
On cross multiplication we get,
 $ \Rightarrow 4 \times y = x$

Taking all the terms to LHS, we get
 $ \Rightarrow x - 4y = 0$ … (2)

To eliminate y we need to multiply \[\left( 1 \right) \times \;2\] , we get
 $ \Rightarrow 10x + 4y = 220$ … (3)

Now, adding (2) and (3), we get
 $ \Rightarrow 10x + 4y + x - 4y = 220 + 0$

On cancelling same terms we get,
 $ \Rightarrow 10x + x = 220$

On adding like terms we get,
 $ \Rightarrow 11x = 220$

On dividing the equation by 11 we get,
 $ \Rightarrow x = \dfrac{{220}}{{11}}$

On simplification we get,
 $ \Rightarrow x = 20$

Now put $x = 20$ in $y = \dfrac{x}{4}$ , we get,
 $ \Rightarrow y = \dfrac{{20}}{4}$

On simplification we get,
 $ \Rightarrow y = 5$

So, we have $x = 20$ and $y = 5$
Hence, solved