NCERT Class 9 Solutions: Linear Equation in Two Variable (Chapter 4) Exercise 4.3 – Part 2

Give line passing equation y-y1=m(x-x1)

Give Line Passing Equation

Give line passing equation y-y1=m(x-x1)

The slope intercept form for the equation of a line

Equation of the Line Y=Mx + C

The slope intercept form for the equation of a line

Q-2 Give the equation of two lines passing through (2,14) . How many more such lines are there, and why?

Solution: For lines passing through (2,14) ; (2, 14) must be a solution. As we will see below there can be infinitely many lines passing through a point (2,14) , with different coefficients of x and y.

  • x=2 and y=14 , one possible relation between these is x+y=2+14=16

    Therefore, x+y=16 is one linear equation passing through point (2, 14) with coefficients of x and y being 1 and 1.

  • Second possible equation can be xy=214=12 (coefficients of x and y are 1 and -1)

    Therefore xy=12orxy+12=0 is another linear equation passing through point (2, 14).

  • Now we can start adding various coefficients to x and y. Let’s construct an equation of form 2x+y=c . Putting the value of the point on the line (2,14) which must satisfy this equation, 2×2+14=c=18 . Our third equation thus becomes, 2x+y=18 .

  • Similarly, 7xy=c is satisfied at (2,14) if c=0

    Therefore, 7xy=0 is another line passing through (2, 14)

    Like we said there can be infinite equations with different coefficients of x and y. It makes sense because through one point infinite lines can pass.

    Infinite lines can pass through a point

    Infinite Line Through a Point

    Infinite lines can pass through a point

Q-3 If the point (3,4) lies on the graph of the equation 3y=ax+7 , find the value of a?

Solution:

Give the point (3,4) lies on the graph of the equation 3y=ax+7

Since point (3,4) lies on the equation, it must satisfy this equation.

Putting the value x=3 and y=4 in the given equation we get,

3×4=a×3+7

12=3a+7

127=3a

5=3a

a=53

Therefore, 3y=ax+7 becomes 3y=53x+7

Check: To check our solutions, we would again substitute, x=3 and y=4

L.H.S 3y=3(4)=12

R.H.S 53x+7=53(3)+7=12

So, L.H.S = R.H.S and hence our solution was correct.

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