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2 years ago

Select the equation of the line that passes through the point (5, 7) and is perpendicular to the line x = 4.

Mathematics

2 answers:

tresset_1 [31]2 years ago

7 0

Solution:

Equation of line perpendicular to , x=4 is

y +k=0

As, keep in mind , the equation of line perpendicular to , ax + by +c=0 is b x - a y +c=0

As, given the line perpendicular to, x =4, passes through the point (5,7).

So, y+k =0, passes through (5,7).

→ 7 +k=0

→ k= -7

So, line perpendicular to x=4, which passes through the point (5,7) is →y-7=0 that is y=7.

Option (C) →y=7

alexandr402 [8]2 years ago

5 0

Answer:

y = 7

Step-by-step explanation:

x=4 is a vertical line that goes through the point x=4. y value of this graph changes while x value will be 4 at every point of the graph.

A line perpendicular to this will be a horizontal line with a fixed y value and the x value will change.

The coordinates of the point of which the line has to pass is given as (5,7). Which means the y coordinate of this line is 7.

As we have established that this line has to have same y value at each point just like x=4 has same x value at each point, we can say the equation of the line is y=7.

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g) $u^{4}\cdot v^{-1}\cdot z^{3}$ , h) $\frac{(x+4)\cdot (x+2)}{3\cdot (x-5)}$

Step-by-step explanation:

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5) $u^{4}\cdot v^{-1}\cdot z^{3}$ $\frac{a^{m}}{a^{n}} = a^{m-n}$ /Result

h) $\frac{x^{2}-16}{x^{2}-10\cdot x + 25} \div \frac{3\cdot x - 12}{x^{2}-3\cdot x -10}$

1) $\frac{x^{2}-16}{x^{2}-10\cdot x + 25} \div \frac{3\cdot x - 12}{x^{2}-3\cdot x -10}$ Given

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6) $\frac{(x+4)\cdot (x+2)}{3\cdot (x-5)}$ $\frac{a^{m}}{a^{n}} = a^{m-n}$ / $\frac{a}{b}\times \frac{c}{d} = \frac{a\cdot c}{b\cdot d}$ /Definition of division/Result