# SSC EXAMS: GRAPHS OF LINEAR EQUATIONS

#### QUERY 1

Number of solutions of two equation 4x – y = 2 and 2y – 8x + 4= 0

A) 1
B) 2
C) Infinitely many solutions
D) 0

MAHA GUPTA
To know the number of solutions of a linear pair of two variables we need to compare the co-efficients of those equations. For this one should re-write the given equations in a similar way. If you see the constant ‘c’ needs to be adjusted as one of them is on the LHS and the other is on the RHS.

So by making them similar they are:
4x – y – 2 = 0, where a1 = 4, b1 = -1 and c1 = -2 —-(i)
-8x + 2y + 4 = 0, where a2 = -8, b2 = 2, and c2 = 4 —-(ii)

Now the rules

1. If a1/a2 ≠ b1/b2 (exactly one solution—-intersecting lines)
2. If a1/a2 = b1/b2 = c1/c2 (infinitely many solutions—-coincident lines)
3. If a1/a2 = b1/b2 ≠ c1/c2 (no solution—-parallel Lines)

In the given set of the equations
a1/a2 = 4/(-8) = -1/2;      b1/b2 = -1/2;      c1/c2 = -2/4 = -1/2

Hence by going through the conditions the pair of the given equations it has INFINITELY MANY SOLUTIONS (option ‘C’)

#### QUERY 2

If a straight line ax+by = c meets x-axis at A and y-axis at B. Then area of triangle OAB where O is the point of intersection of coordinate axes, is?

A) b²/2ac
B) a²/2bc
C) c²/2ab
D) a²b²/2c

MAHA GUPTA
A triangle formed by any points on x-axis and y-axis with the point of their intersection is always a right angle, in which the angle on the intersection is a right angle. We also know that area of a right triangle = multiplication of its sides forming the right angle/2

According to above ΔOAB is a right angle triangle where angle AOB = 90°; and its area = (OA*OB)/2

Now we are left with finding the lengths of sides OA and OB

At point A, y = 0, so x = c/a; means its coordinates are (c/a, 0); means OA = c/a
And at point B, x = 0, so y = c/b; means its coordinates are (0, c/b); means OB = c/b

Hence the area of OAB = [(c/a)*(c/b)]/2
= c²/2ab (option ‘C’)

#### QUERY 3

The length of the part of the graph of the equation 4x + 3y – 12 = 0 intercepted by two co-ordinate axes is?

A) 7 units
B) 5 units
C) 6 units
D) 2.5 units

MAHA GUPTA
The part of the graph i.e. segment of the line of the graph is intercepted by the two co-ordinates, means one of the point of the ordinates is meeting axes-X is y = 0 and one of the point of the ordinates is meeting axes-Y is x = 0.

Therefore you have just to find the corresponding value of y when x = 0, and corresponding value of x when y = 0

Thus, putting x = 0 in the above equation, y = 4. and putting y = 0, x = 3

So, the two of the coordinating pairs are (3, 0) lying on X-axes, and (0, 4) lying on Y-axis

We know that graph of any equation intercepted by two co-ordinate axes makes a right triangle, the length of the part of the graph intercepted by two co-ordinate being the hypotenuse of that triangle.

Now, length of the side of the triangle on X-axis = 3 units                   —-co-ordinates of point (3, 0) is giving that

And length of the side of the triangle on Y-axis = 4 units                      —–co-ordinates of point (0, 4) is giving that

Hence by Pythagoras = The length of the part of the graph of the equation intercepted by two co-ordinate axes is i.e. hypotenuse of the triangle = √(3² + 4²) = 5 units (option ‘B’)

#### QUERY 4

The value of k for which the graph of (k – 1)x + y – 2 = 0 and (2 – k)x – 3y + 1 = 0 are parallel is?

A) 1/2
B) -1/2
C) 2
D) -2

RONNIE BANSAL
For the graph of two equations to be parallel
a1/a2 = b1/b2 ≠ c1/c2; where a & b are the coefficients of x & y respectively and c is the constant term.

So (k -1)/(2 -k) = 1/-3

=> (k – 1)/1 = (2 – k)/-3

=> k = 1/2 (option ‘A’)

We see that it’s not equal to the ratio of the constant terms i.e. -2/1; hence for k = 1/2 the graph of the given equations is parallel.

#### QUERY 5

The value of p for which the graph of 4x + py + 8 = 0 and 2x + 2y + 2 = 0 has unique solution?

A) 4
B) except 4
C) except 3
D) except 2

MAHA GUPTA
Condition for unique solution:
a1a≠ b1b2

Here a1 = 4,    a= 2,     b= p,     b= 2

Now for the given pair to have a unique solution:
4/2 ≠ p/2

=> p ≠ 4

Therefore, for all values of p, except 4, the given pair of equations will have a unique solution. (option ‘B’)

#### QUERY 6

The value of k for which the graph of kx + 3y – (k – 3) = 0 and 12x + ky – k = 0 has infinitely many solutions?

A) 6
B) -6
C) 0
D) ±6

MAHA GUPTA
For a pair of linear equations to have infinitely many solutions:
a1/a2 = b1/b2 = c1/c2

Here, a1/a2 = k/12      b1/b2 = 3/k       c1/c= (k – 3)/k

So, k/12 = 3/k          and          3/k = (k – 3)/k
=> k² = 36                and          k² = 6k
=> k = ±6                  and          k = 0 or k = 6

Therefore, the value of k, that satisfies both the conditions, is k = 6. For this value, the pair of linear equations has infinitely many solutions (option ‘A’)

#### QUERY 7

For which values of p and q, will the following pair of linear equations have infinitely many solutions?
4x + 5 y = 2
(2p + 7q)x + (p + 8q)y = 2q – p + 1.

A) 1,   2
B) 2,   -1
C) -1,  2
D) 3,    -1

MAHA GUPTA
For a pair of linear equations to have infinitely many solutions:
a1/a2 = b1/b2 = c1/c2

Here, a1/a242p + 7q
b1/b2 = 5p + 8q
c1/c22q – p + 1

S0, 42p + 7q = 5p + 8q 22q – p + 1

=> 42p + 7q = 5p + 8q          and          5p + 8q 22q – p + 1

=> q = –2p      —- (i)      and          4p + 3q = 2         —- (ii)

Solving (i) & (ii)
p = -1; q = 2
So, for p = –1, q = 2, the given pair of linear equations will have infinitely many solutions (option ‘C’)

#### QUERY 8

Find the area of the figure formed by the graphs of the lines x = –2 and y = 3 of the x-axis and the y-axis.

A) 5.5 square units
B) 6 square units
C) 6.5 square units
D) 9 square units

MAHA GUPTA
We know that the graph of x = –2 is a line parallel to y-axis at a distance of 2 units to the left of it.
The graph of y = 3 is a line parallel to the x-axis at a distance of 3 units above it.

You’ll see, thus, the figure enclosed by the lines x = –2, y = 3 will be a rectangle with 2 units as its breadth and 3 units as its length.

As the area of a rectangle = length × breadth,
Therefore area of the figure so formed = 3 × 2 = 6 square units (option ‘B’)

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