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olya-2409 [2.1K]
3 years ago
9

Help please thank you !!

Mathematics
1 answer:
swat323 years ago
6 0

Answer:

AC and AE

Step-by-step explanation:

Those are the only secants in the image shown.

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What is the equation of the line that passes through the point (1,0) and is perpendicular to the line x+5y=30
Andre45 [30]

Answer:

The equation of line passing through points (1 , 0) is  x - 5 y - 1 = 0    

Step-by-step explanation:

Given equation of line as

x + 5 y = 30

Now, equation of line in standard form is y = m x + c

where m is the slope

So,  x + 5 y = 30

Or, 5 y = - x + 30

Or, y = - \frac{1}{5} x + 6

So, Slope of this line m = - \frac{1}{5}

Again , let the slope of other line passing through point (1 , 0) is M

And Both lines are perpendicular , So , products of line = - 1

i.e m × M = - 1

Or, M = - \frac{1}{m}

Or, M = - 1 × - \frac{1}{\frac{1}{5}} =  \frac{1}{5}

So, equation of line with slope M and points (1, 0) is

y - y_1 = M × (x - x_1)

Or, y - ( 0 ) =  \frac{1}{5} × ( x - 1 )

Or, y  =  \frac{1}{5} x - \frac{1}{5} × 1

Or, y  =   \frac{1}{5} x - \frac{1}{5}

or, y + \frac{1}{5} =   \frac{1}{5} x

Or, 5×y + 1  =   x

∴ 5 y + 1 =  x

I.e  x - 5 y - 1 = 0

Hence The equation of line passing through points (1 , 0) is  x - 5 y - 1 = 0   Answer

7 0
3 years ago
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