Question

the line y=mx+c is parallel to the line y=2x+4. the distance AB is 6 units. find the value of m and c

Answer 1

SOLUTION:-

y = mx + c

mx - y + c = 0 ......(1)

a = m, b = -1, c = c

y = 2x + 4

2x - y + 4 = 0 .......(2)

a = 2, b = -1, c = 4

these lines are parallel to each other.

•°• m = 2 ....... {°•° The coefficient of x and y in both equation to be the same}

Now,

$</p><p></p><p>\tt \: p = \left | </p><p> \begin{array}{c c}</p><p></p><p> (c1 & -c2) \div</p><p> \sqrt{a^2 + b^2}</p><p> </p><p> \end {array}</p><p> \right |</p><p></p><p>$

now,

6 = (c - 4)/ √( 2² + (-1)²)

6 = (c - 4)/ √( 2² + 1²)

6 = (c - 4)/ √(4 + 1)

6 = (c - 4)/ √(5)

6√(5) = c - 4

c = 4 + 6√(5)

The value for m is 2 and for c is 4 + 6√(5)

Answer 2

Given ,

The two lines are parallel to each other

• y = mx + c
• y = 2x + 4

And the distance between them is 6 units

We know that ,

If two lines are parallel to each other then their slopes are equal

Thus ,

m = 2

Now ,

The distance between two parrallel lines is given by

$\large \sf \fbox{D = \frac{ | C_{1} -C_{2} | }{ \sqrt{ {(A)}^{2} + {(B)}^{2} } } }$

Thus ,

$\sf \mapsto 6 = \frac{ |c -4 | }{ \sqrt{{(2)}^{2} + {( - 1)}^{2}} } \\ \\\sf \mapsto 6 = \frac{ ±</p><p>(c - 4) }{ \sqrt{5} } \\ \\\sf \mapsto c = 6 \sqrt{5} - 4 \: \: or \: \: c = 6 \sqrt{5} + 4$

$\sf \underline{The \: value \: of \: m \: and \: c \: are \: 2 \: and \: 6 \sqrt{5} - 4 \: or \: 6 \sqrt{5} + 4}$