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goldfiish [28.3K]
3 years ago
15

I need the answer for this exact question

Mathematics
1 answer:
denis23 [38]3 years ago
3 0

Answer:

y=\frac{7}{2}x-20

Step-by-step explanation:

Let the equation of the line be y-y_1=m(x-x_1) where, 'm' is its slope and (x_1,y_1) is a point on it.

Given:

The equation of a known line is:

y=-\frac{2}{7}x+9

A point on the unknown line is:

(x_1,y_1)=(4,-6)

Both the lines are perpendicular to each other.

Now, the slope of the known line is given by the coefficient of 'x'. Therefore, the slope of the known line is m_1=-\frac{2}{7}

When two lines are perpendicular, the product of their slopes is equal to -1.

Therefore,

m\cdot m_1=-1\\m=-\frac{1}{m_1}\\m=-\frac{1}{\frac{-2}{7} }=\frac{7}{2}

Therefore, the equation of the unknown line is determined by plugging in all the given values. This gives,

y-(-6))=\frac{7}{2}(x-4)\\y+6=\frac{7}{2}x-14\\y=\frac{7}{2}x-14-6\\\\y=\frac{7}{2}x-20

The equation of a line perpendicular to the given line and passing through (4, -6) is y=\frac{7}{2}x-20.

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ivann1987 [24]

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At Rose State College Allison Wells received the following grades her accounting class 93, 60, 78, 89 ,78 and 82 Allison's instr
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Two isosceles triangles share the same base. Prove that the medians to this base are collinear. (There are two cases in this pro
Anna71 [15]

Answer:

Given: Two Isosceles triangle ABC and Δ PBC having same base BC. AD is the median of Δ ABC and PD is the median of Δ PBC.

To prove: Point A,D,P are collinear.

Proof:

→Case 1.  When vertices A and P are opposite side of Base BC.

In Δ ABD and Δ ACD

AB= AC   [Given]

AD is common.

BD=DC  [median of a triangle divides the side in two equal parts]

Δ ABD ≅Δ ACD [SSS]

∠1=∠2 [CPCT].........................(1)

Similarly, Δ PBD ≅ Δ PCD [By SSS]

 ∠ 3 = ∠4 [CPCT].................(2)

But,  ∠1+∠2+ ∠ 3 + ∠4 =360° [At a point angle formed is 360°]

2 ∠2 + 2∠ 4=360° [using (1) and (2)]

∠2 + ∠ 4=180°

But ,∠2 and ∠ 4 forms a linear pair i.e Point D is common point of intersection of median AD and PD of ΔABC and ΔPBC respectively.

So, point A, D, P lies on a line.

CASE 2.

When ΔABC and ΔPBC lie on same side of Base BC.

In ΔPBD and ΔPCD

PB=PC[given]

PD is common.

BD =DC [Median of a triangle divides the side in two equal parts]

ΔPBD ≅ ΔPCD  [SSS]

∠PDB=∠PDC [CPCT]

Similarly, By proving ΔADB≅ΔADC we will get,  ∠ADB=∠ADC[CPCT]

As PD and AD are medians to same base BC of ΔPBC and ΔABC.

∴ P,A,D lie on a line i.e they are collinear.




3 0
3 years ago
A-2b= -10 for b plsssss helpppp
Artyom0805 [142]
A-2b=-10
b=-10-a.
Hope this is the answer
Mark as brainliest
5 0
3 years ago
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